{"question_id": 562694, "answer_id": 563063, "title": "Integral $\\int_{-1}^1\\frac1x\\sqrt{\\frac{1+x}{1-x}}\\ln\\left(\\frac{2\\,x^2+2\\,x+1}{2\\,x^2-2\\,x+1}\\right) \\mathrm dx$", "url": "https://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1", "tags": ["calculus", "integration", "definite-integrals", "contour-integration", "closed-form"], "asker": "Laila Podlesny", "question_score": 638, "answer_score": 480, "answer_date": "2013-11-11", "type": "integral", "question_markdown": "I need help with this integral:\n\n$$I=\\int_{-1}^1\\frac1x\\sqrt{\\frac{1+x}{1-x}}\\ln\\left(\\frac{2x^2+2x+1}{2x^2-2x+1}\\right)\\mathrm{d}x$$\n\nThe integrand graph looks like this:\n\n$\\hspace{1in}$![The integrand graph][1]\n\nThe approximate numeric value of the integral:\n$$I\\approx8.372211626601275661625747121...$$\n\nNeither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in [WolframAlpha](http://www.wolframalpha.com) and [ISC+](http://isc.carma.newcastle.edu.au) did not return plausible closed form candidates either. But I still hope there might be a closed form for it. \n\nI am also interested in cases when only the numerator or only the denominator is present under the logarithm.\n\n[1]: https://i.sstatic.net/ztodM.png", "cleo_answer_markdown": "$\\large\\hspace{3in}I=4\\,\\pi\\operatorname{arccot}$[$\\sqrt\\phi$](https://mathworld.wolfram.com/GoldenRatio.html)", "n": 1, "slug": "q562694-integral-int-1-1-frac1x-sqrt-frac-1-x-1-x-ln-lef"}
{"question_id": 712798, "answer_id": 713518, "title": "Crazy $\\int_0^\\infty{_3F_2}\\left(\\begin{array}c\\tfrac58,\\tfrac58,\\tfrac98\\\\\\tfrac12,\\tfrac{13}8\\end{array}\\middle|\\ {-x}\\right)^2\\frac{dx}{\\sqrt x}$", "url": "https://math.stackexchange.com/questions/712798/crazy-int-0-infty-3f-2-left-beginarrayc-tfrac58-tfrac58-tfrac98-tfra", "tags": ["calculus", "integration", "definite-integrals", "closed-form", "hypergeometric-function"], "asker": "Zakharia Stanley", "question_score": 26, "answer_score": 130, "answer_date": "2014-03-15", "type": "integral", "question_markdown": "Is there any chance to find a closed form for this integral?\n$$I=\\int_0^\\infty{_3F_2}\\left(\\begin{array}c\\tfrac58,\\tfrac58,\\tfrac98\\\\\\tfrac12,\\tfrac{13}8\\end{array}\\middle|\\ {-x}\\right)^2\\frac{dx}{\\sqrt x}$$", "cleo_answer_markdown": "$$I=\\frac{50\\,\\pi^{3/2}}{3\\,\\Gamma^2\\left(\\frac14\\right)}\\Big(\\ln\\left(3+\\sqrt8\\right)-\\sqrt2\\Big)$$", "n": 2, "slug": "q712798-crazy-int-0-infty-3f-2-left-begin-array-c-tfrac5"}
{"question_id": 908108, "answer_id": 908325, "title": "How to find ${\\large\\int}_0^1\\frac{\\ln^3(1+x)\\ln x}x\\mathrm dx$", "url": "https://math.stackexchange.com/questions/908108/how-to-find-large-int-01-frac-ln31x-ln-xx-mathrm-dx", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "harmonic-numbers"], "asker": "Oksana Gimmel", "question_score": 119, "answer_score": 95, "answer_date": "2014-08-25", "type": "integral", "question_markdown": "Please help me to find a closed form for this integral:\n$$I=\\int_0^1\\frac{\\ln^3(1+x)\\ln x}x\\mathrm dx\\tag1$$\nI suspect it might exist because there are similar integrals having closed forms:\n$$\\begin{align}\\int_0^1\\frac{\\ln^3(1-x)\\ln x}x\\mathrm dx&=12\\zeta(5)-\\pi^2\\zeta(3)\\tag2\\\\\n\\int_0^1\\frac{\\ln^2(1+x)\\ln x}x\\mathrm dx&=\\frac{\\pi^4}{24}-\\frac16\\ln^42+\\frac{\\pi^2}6\\ln^22-\\frac72\\zeta(3)\\ln2-4\\operatorname{Li}_4\\!\\left(\\tfrac12\\right)\\tag3\\\\\n\\int_0^1\\frac{\\ln^3(1+x)\\ln x}{x^2}\\mathrm dx&=\\frac34\\zeta(3)-\\frac{63}4\\zeta(3)\\ln2+\\frac{23\\pi^4}{120}\\\\&-\\frac34\\ln^42-2\\ln^32+\\frac{3\\pi^2}4\\ln^22-18\\operatorname{Li}_4\\!\\left(\\tfrac12\\right).\\tag4\\end{align}$$\nThanks!", "cleo_answer_markdown": "Indeed, there is a closed form for this integral:\n$$I=\\frac{\\pi^2}3\\ln^32-\\frac25\\ln^52+\\frac{\\pi^2}2\\zeta(3)+\\frac{99}{16}\\zeta(5)-\\frac{21}4\\zeta(3)\\ln^22\\\\-12\\operatorname{Li}_4\\left(\\frac12\\right)\\ln2-12\\operatorname{Li}_5\\left(\\frac12\\right).$$", "n": 3, "slug": "q908108-how-to-find-large-int-0-1-frac-ln-3-1-x-ln-x-x-m"}
{"question_id": 1142705, "answer_id": 1154092, "title": "Evaluate $\\int_0^{\\pi/2}\\frac{x^2\\log^2{(\\sin{x})}}{\\sin^2x}dx$", "url": "https://math.stackexchange.com/questions/1142705/evaluate-int-0-pi-2-fracx2-log2-sinx-sin2xdx", "tags": ["definite-integrals", "improper-integrals"], "asker": "user178256", "question_score": 30, "answer_score": 78, "answer_date": "2015-02-18", "type": "integral", "question_markdown": "How  evaluate this integral?\n $$I=\\int_0^{\\pi/2}\\frac{x^2\\log^2{(\\sin{x})}}{\\sin^2x}\\,dx$$\n Note: $$\\int_0^{\\pi/2}\\frac{x^2\\log{(\\sin x)}}{\\sin^2x}dx=\\pi\\ln{2}-\\frac{\\pi}{2}\\ln^22-\\frac{\\pi^3}{12}.$$", "cleo_answer_markdown": "$$I=\\frac\\pi3\\ln^32-\\pi\\ln^22+2\\pi\\ln2+\\frac{\\pi^3}6\\left(\\ln2-1\\right)+\\frac\\pi8\\zeta(3)$$", "n": 4, "slug": "q1142705-evaluate-int-0-pi-2-frac-x-2-log-2-sin-x-sin-2x"}
{"question_id": 1595389, "answer_id": 1595449, "title": "Integral ${\\large\\int}_0^{\\pi/2}\\arctan^2\\!\\left(\\frac{\\sin x}{\\sqrt3+\\cos x}\\right)dx$", "url": "https://math.stackexchange.com/questions/1595389/integral-large-int-0-pi-2-arctan2-left-frac-sin-x-sqrt3-cos-x-ri", "tags": ["calculus", "integration", "trigonometry", "definite-integrals", "closed-form"], "asker": "Oksana Gimmel", "question_score": 20, "answer_score": 70, "answer_date": "2015-12-31", "type": "integral", "question_markdown": "I need to evaluate this integral:\n$$I=\\int_0^{\\pi/2}\\arctan^2\\!\\left(\\frac{\\sin x}{\\sqrt3+\\cos x}\\right)dx$$\n_Maple_ and _Mathematica_ cannot evaluate it in this form.\nIts numeric value is\n$$I\\approx0.156371391375711701230837603266631522020409597791339398428...$$\nthat is not recognized by [_WolframAlpha_](http://www.wolframalpha.com/) and [_Inverse Symbolic Calculator+_](http://isc.carma.newcastle.edu.au/).\n\nIs it possible to evaluate this integral in a closed form?\n\nI found similar questions [here](https://math.stackexchange.com/q/1593334/75621), [here](https://math.stackexchange.com/q/564816/75621) and [here](https://math.stackexchange.com/q/1048045/75621), but approaches shown in the answers do not seem to be directly applicable here.", "cleo_answer_markdown": "$$I=\\frac\\pi{20}\\ln^23+\\frac\\pi4\\operatorname{Li_2}\\left(\\tfrac13\\right)-\\frac15\\operatorname{Ti}_3\\left(\\sqrt3\\right),$$\nwhere\n$$\\operatorname{Ti}_3\\left(\\sqrt3\\right)=\\Im\\Big[\\operatorname{Li}_3\\left(i\\sqrt3\\right)\\Big]=\\frac{\\sqrt3}8\\Phi\\left(-3,3,\\tfrac12\\right)=\\frac{5\\sqrt3}4\\,{_4F_3}\\!\\left(\\begin{array}c\\tfrac12,\\tfrac12,\\tfrac12,\\tfrac12\\\\\\tfrac32,\\tfrac32,\\tfrac32\\end{array}\\middle|\\tfrac34\\right)-\\frac{5\\pi^3}{432}-\\frac\\pi{16}\\ln^23.$$", "n": 5, "slug": "q1595389-integral-large-int-0-pi-2-arctan-2-left-frac-sin"}
{"question_id": 1588996, "answer_id": 1589052, "title": "Yet another log-sin integral $\\int\\limits_0^{\\pi/3}\\log(1+\\sin x)\\log(1-\\sin x)\\,dx$", "url": "https://math.stackexchange.com/questions/1588996/yet-another-log-sin-integral-int-limits-0-pi-3-log1-sin-x-log1-sin-x", "tags": ["integration", "trigonometry", "definite-integrals", "logarithms", "polylogarithm"], "asker": "X.C.", "question_score": 29, "answer_score": 61, "answer_date": "2015-12-25", "type": "integral", "question_markdown": "There has been much interest to various log-trig integrals on this site (e.g. see [[1]](https://math.stackexchange.com/q/553297/79756)[[2]](https://math.stackexchange.com/q/699746/79756)[[3]](https://math.stackexchange.com/q/751375/79756)[[4]](https://math.stackexchange.com/q/771610/79756)[[5]](https://math.stackexchange.com/q/775200/79756)[[6]](https://math.stackexchange.com/q/775206/79756)[[7]](https://math.stackexchange.com/q/784508/79756)[[8]](https://math.stackexchange.com/q/798227/79756)[[9]](https://math.stackexchange.com/q/1164183/79756)).\nHere is another one I'm trying to solve:\n$$\\int\\limits_0^{\\pi/3}\\log(1+\\sin x)\\log(1-\\sin x)\\,dx\\approx-0.41142425522824105371...$$\nI tried to feed it to _Maple_ and _Mathematica_, but they are unable to evaluate in this form. After changing the variable $x=2\\arctan z,$ and factoring rational functions under logarithms, the integrand takes the form\n$$\\frac{2 \\log ^2\\left(z^2+1\\right)}{z^2+1}-\\frac{4 \\log (1-z) \\log \\left(z^2+1\\right)}{z^2+1}\\\\-\\frac{4 \\log (z+1) \\log\n   \\left(z^2+1\\right)}{z^2+1}+\\frac{8 \\log (1-z) \\log (z+1)}{z^2+1}$$\nin which it can be evaluated by _Mathematica_. It spits out a huge ugly expression with complex numbers, polylogarithms, polygammas and generalized hypergeometric functions (that indeed matches numerical estimates of the integral). It takes a long time to simplify and with only little improvement (see [here](https://gist.githubusercontent.com/anonymous/37bebbfa3c2bdbfa4c83/raw/bf38352f660364c301ed062fad50488254669c81/gistfile1.txt) if you are curious). \n\nI'm looking for a better approach to this integral that can produce the answer in a simpler form.", "cleo_answer_markdown": "$$\\begin{align}\\int_0^{\\pi/3}\\ln(1+\\sin x)\\ln(1-\\sin x)\\,dx=&\\frac{29\\pi^3}{216}+\\frac{5\\pi}6\\ln^2\\left(2+\\sqrt3\\right)+\\frac\\pi3\\ln^22+\\frac{\\pi^2}{3\\sqrt3}\\ln2\\\\+&\\frac{8G}3\\ln\\left(2+\\sqrt3\\right)-4 \\operatorname{Ti}_3\\left(2+\\sqrt3\\right)-\\frac{\\psi^{(1)}\\!\\left(\\tfrac13\\right)}{2 \\sqrt{3}}\\ln2,\\end{align}$$\nwhere $G$ is the Catalan constant, $\\operatorname{Ti}_3(z)=\\Im\\operatorname{Li}_3(iz)$ is the generalized inverse tangent integral, and $\\psi^{(1)}(z)$ is the trigamma function.", "n": 6, "slug": "q1588996-yet-another-log-sin-integral-int-limits-0-pi-3-l"}
{"question_id": 418134, "answer_id": 582370, "title": "Calculating $\\int_{\\pi/2}^{\\pi}\\frac{x\\sin{x}}{5-4\\cos{x}}\\,\\mathrm dx$", "url": "https://math.stackexchange.com/questions/418134/calculating-int-pi-2-pi-fracx-sinx5-4-cosx-mathrm-dx", "tags": ["calculus", "integration", "definite-integrals"], "asker": "YMJou", "question_score": 29, "answer_score": 57, "answer_date": "2013-11-26", "type": "integral", "question_markdown": "> Calculate the following integral:$$\\int_{\\pi/2}^{\\pi}\\frac{x\\sin{x}}{5-4\\cos{x}}\\,\\mathrm dx$$\n\nI can calculate the integral on $[0,\\pi]$,but I want to know how to do it on $[\\frac{\\pi}{2},\\pi]$.", "cleo_answer_markdown": "$$\\begin{align}\\int_{\\pi/2}^\\pi\\frac{x\\sin x}{5-4\\cos x}dx&=\\pi\\left(\\frac{\\ln3}2-\\frac{\\ln2}4-\\frac{\\ln5}8\\right)-\\frac12\\operatorname{Ti}_2\\left(\\frac12\\right)\\\\&=\\pi\\left(\\frac{\\ln3}2-\\frac{\\ln2}4-\\frac{\\ln5}8\\right)-\\frac12\\Im\\,\\chi_2\\left(\\frac{\\sqrt{-1}}2\\right),\\end{align}$$\nwhere  $\\operatorname{Ti}_2(z)$ is the [inverse tangent integral](http://mathworld.wolfram.com/InverseTangentIntegral.html) and $\\Im\\,\\chi_\\nu(z)$ is the [imaginary part](http://mathworld.wolfram.com/ImaginaryPart.html) of the [Legendre chi function](http://mathworld.wolfram.com/LegendresChi-Function.html).\n\n---\n_Hint:_\nUse the following Fourier series and integrate termwise:\n$$\\frac{\\sin x}{5-4\\cos x}=\\sum_{n=1}^\\infty\\frac{\\sin n x}{2^{n+1}}.$$", "n": 7, "slug": "q418134-calculating-int-pi-2-pi-frac-x-sin-x-5-4-cos-x-m"}
{"question_id": 1279165, "answer_id": 1279591, "title": "Integrals of the form ${\\large\\int}_0^\\infty\\operatorname{arccot}(x)\\cdot\\operatorname{arccot}(a\\,x)\\cdot\\operatorname{arccot}(b\\,x)\\ dx$", "url": "https://math.stackexchange.com/questions/1279165/integrals-of-the-form-large-int-0-infty-operatornamearccotx-cdot-operat", "tags": ["calculus", "integration", "trigonometry", "definite-integrals", "logarithms"], "asker": "Vladimir Reshetnikov", "question_score": 68, "answer_score": 54, "answer_date": "2015-05-12", "type": "integral", "question_markdown": "I'm interested in integrals of the form\n$$I(a,b)=\\int_0^\\infty\\operatorname{arccot}(x)\\cdot\\operatorname{arccot}(a\\,x)\\cdot\\operatorname{arccot}(b\\,x)\\ dx,\\color{#808080}{\\text{ for }a>0,\\,b>0}\\tag1$$\nIt's known$\\require{action}\\require{enclose}\\texttip{{}^\\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)}$ that\n$$I(a,0)=\\frac{\\pi^2}4\\left[\\ln\\left(1+\\frac1a\\right)+\\frac{\\ln(1+a)}a\\right].\\tag2$$\n_Maple_ and _Mathematica_ are also able to evaluate\n$$I(1,1)=\\frac{3\\pi^2}4\\ln2-\\frac{21}8\\zeta(3).\\tag3$$\n\n---\nIs it possible to find a general closed form for $I(a,1)$? Or, at least, for $I(2,1)$ or $I(3,1)$?", "cleo_answer_markdown": "$$\\begin{align}I(2,1)&=\\frac{\\pi^2}3\\ln2-\\frac{\\pi^2}6\\ln3+2\\ln^22\\cdot\\ln3-3\\ln2\\cdot\\ln^23+\\frac{29}{24}\\ln^33\\\\&+\\frac{73}{16}\\zeta(3)-2\\ln2\\cdot\\operatorname{Li}_2\\left(\\tfrac13\\right)-\\frac{13}4\\operatorname{Li}_3\\left(\\tfrac13\\right)-4\\operatorname{Li}_3\\left(\\tfrac23\\right)\\end{align}$$\n\n---\n$$\\begin{align}I(3,1)&=\\frac{13\\,\\pi^2}{12}\\ln2-\\frac{4\\,\\pi^2}9\\ln3-\\frac13\\ln2\\cdot\\ln^23+\\frac7{18}\\ln^33\\\\&-\\frac{13}8\\zeta(3)+\\ln3\\cdot\\operatorname{Li}_2\\left(\\tfrac13\\right)+\\frac43\\operatorname{Li}_3\\left(\\tfrac13\\right)-\\frac23\\operatorname{Li}_3\\left(\\tfrac23\\right)\\end{align}$$\n\n---\nUpdate (in response to a comment):\n$$\\begin{align}&I(\\phi,1)=\\frac32\\ln^32+\\frac{\\pi^2}{12}\\Big[\\left(6-3\\sqrt5\\right)\\ln2+\\left(3\\sqrt5-4\\right)\\ln\\left(1+\\sqrt5\\right)\\Big]+\\frac{51-21\\sqrt5}{48}\\zeta(3)\\\\&-\\frac{\\ln\\left(1+\\sqrt5\\right)}2\\Bigg[15\\ln^22-15\\ln\\left(1+\\sqrt5\\right)\\ln2+4\\ln^2\\left(1+\\sqrt5\\right)+2\\operatorname{Li}_2\\left(\\frac{1-\\sqrt5}4\\right)\\Bigg]\\\\&-\\ln\\left(3+\\sqrt5\\right)\\operatorname{Li}_2\\left(\\sqrt5-2\\right)+\\frac{11+3\\sqrt5}{48}\\operatorname{Li}_3\\left(9-4\\sqrt5\\right)-\\frac{13+3\\sqrt5}6\\operatorname{Li}_3\\left(\\sqrt5-2\\right)\\end{align}$$", "n": 8, "slug": "q1279165-integrals-of-the-form-large-int-0-infty-operator"}
{"question_id": 905653, "answer_id": 905723, "title": "How to find ${\\large\\int}_1^\\infty\\frac{1-x+\\ln x}{x \\left(1+x^2\\right) \\ln^2 x} \\mathrm dx$?", "url": "https://math.stackexchange.com/questions/905653/how-to-find-large-int-1-infty-frac1-x-ln-xx-left1x2-right-ln2-x", "tags": ["calculus", "integration", "definite-integrals", "logarithms", "closed-form"], "asker": "Laila Podlesny", "question_score": 28, "answer_score": 50, "answer_date": "2014-08-22", "type": "integral", "question_markdown": "Please help me to find a closed form for this integral:\n$$\nI=\\int_1^\\infty\\frac{1-x+\\ln x}{x \\left(1+x^2\\right) \\ln^2 x} \\mathrm dx\n$$\n\n\nRoutine textbook methods for this complicated integral fail.", "cleo_answer_markdown": "$$I=\\frac{\\ln2}6+\\frac{\\ln\\pi}2-6\\ln A+\\frac2\\pi G,$$\nwhere $A$ is the [Glaisher–Kinkelin constant](http://en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant) and $G$ is the [Catalan constant](http://en.wikipedia.org/wiki/Catalan's_constant).", "n": 9, "slug": "q905653-how-to-find-large-int-1-infty-frac-1-x-ln-x-x-le"}
{"question_id": 714628, "answer_id": 953505, "title": "Closed form for $\\int_{-\\infty}^0\\operatorname{Ei}^3x\\,dx$", "url": "https://math.stackexchange.com/questions/714628/closed-form-for-int-infty0-operatornameei3x-dx", "tags": ["calculus", "integration", "definite-integrals", "special-functions", "closed-form"], "asker": "Vladimir Reshetnikov", "question_score": 21, "answer_score": 47, "answer_date": "2014-10-01", "type": "integral", "question_markdown": "Let $\\operatorname{Ei}x$ denote the [exponential integral](http://mathworld.wolfram.com/ExponentialIntegral.html):\n$$\\operatorname{Ei}x=-\\int_{-x}^\\infty\\frac{e^{-t}}tdt.\\tag1$$\nIt's not difficult to find that\n$$\\int\\operatorname{Ei}x\\,dx=x\\,\\operatorname{Ei}x-e^x,\\tag2$$\n$$\\int\\operatorname{Ei}^2x\\,dx=x\\,\\operatorname{Ei}^2x-2\\,e^x\\operatorname{Ei}x+2\\,\\operatorname{Ei}(2x)\\tag3$$\nand\n$$\\int_{-\\infty}^0\\operatorname{Ei}x\\,dx=-1,\\tag4$$\n$$\\int_{-\\infty}^0\\operatorname{Ei}^2x\\,dx=\\ln4.\\tag5$$\n\n---\nIs it possible generalize these results for higher powers of $\\operatorname{Ei}x$?  \nIn particular, are there closed forms for\n$$\\int\\operatorname{Ei}^3x\\,dx\\tag6$$\nor\n$$\\int_{-\\infty}^0\\operatorname{Ei}^3x\\,dx\\ ?\\tag7$$", "cleo_answer_markdown": "$$\\begin{align}\\int_0^\\infty\\operatorname{Ei}^3(-x)\\,dx&=-3\\operatorname{Li}_2\\left(\\frac14\\right)-6\\ln^22.\\\\\\\\\\int_0^\\infty\\operatorname{Ei}^4(-x)\\,dx&=24\\operatorname{Li}_3\\left(\\frac14\\right)-48\\operatorname{Li}_2\\left(\\frac13\\right)\\ln2-13\\,\\zeta(3)\\\\&-32\\ln^32+48\\ln^22\\,\\ln3-24\\ln2\\,\\ln^23+6\\pi^2\\ln2.\\end{align}$$", "n": 10, "slug": "q714628-closed-form-for-int-infty-0-operatorname-ei-3x-d"}
{"question_id": 918680, "answer_id": 920995, "title": "Closed Form for the Imaginary Part of $\\text{Li}_3\\Big(\\frac{1+i}2\\Big)$", "url": "https://math.stackexchange.com/questions/918680/closed-form-for-the-imaginary-part-of-textli-3-big-frac1i2-big", "tags": ["integration", "closed-form", "polylogarithm"], "asker": "Lucian", "question_score": 49, "answer_score": 43, "answer_date": "2014-09-06", "type": "other", "question_markdown": ">$\\qquad\\qquad$ *Is there any closed form expression for the imaginary part of* $~\\text{Li}_3\\bigg(\\dfrac{1+i}2\\bigg)$ ?\n\n___\n*Motivation: We [already](https://math.stackexchange.com/questions/918135) [know](https://math.stackexchange.com/questions/917154) that* $~\\Re\\bigg[\\text{Li}_3\\bigg(\\dfrac{1+i}2\\bigg)\\bigg]=\\dfrac{\\ln^32}{48}-\\dfrac5{192}~\\pi^2~\\ln2+\\dfrac{35}{64}~\\zeta(3)$, \n\n*so I was wondering whether a similar closed form expression might also exist for its*\n\n*imaginary part as well. Thank you !*\n___\n[*Apparently*](http://www.arxiv.org/pdf/1911.12155.pdf), $~\\Im~\\text{Li}_3\\bigg(\\dfrac{1+i}2\\bigg)~+~\\Im~\\text{Li}_3(1+i)~=~\\dfrac7{128}\\cdot\\pi^3~+~\\dfrac3{32}\\cdot\\pi\\cdot\\ln^22,~$ *so the question*\n\n*is equivalent to asking for the closed form of the imaginary part of* $~\\Im~\\text{Li}_3(1+i).$", "cleo_answer_markdown": "If you consider a hypergeometric function to be a closed form, you can have the following result:\n$$\\Im\\left[\\operatorname{Li}_3\\left(\\frac{1+i}2\\right)\\right]=\\frac{\\pi^3}{128}+\\frac\\pi{32}\\ln^22+\\frac14\\,{_4F_3}\\left(\\begin{array}c\\tfrac12,\\tfrac12,1,1\\\\\\tfrac32,\\tfrac32,\\tfrac32\\end{array}\\middle|\\,1\\right).$$\nAnd for the polylogarithm value that appears in another answer, you can have\n$$\\Im\\Big[\\operatorname{Li}_3\\left(1+i\\right)\\Big]=\\frac{3\\pi^3}{64}+\\frac\\pi{16}\\ln^22-\\frac14\\,{_4F_3}\\left(\\begin{array}c\\tfrac12,\\tfrac12,1,1\\\\\\tfrac32,\\tfrac32,\\tfrac32\\end{array}\\middle|\\,1\\right).$$", "n": 11, "slug": "q918680-closed-form-for-the-imaginary-part-of-text-li-3"}
{"question_id": 918821, "answer_id": 919831, "title": "Closed form for ${\\large\\int}_0^1\\frac{\\ln^2x}{\\sqrt{1-x+x^2}}dx$", "url": "https://math.stackexchange.com/questions/918821/closed-form-for-large-int-01-frac-ln2x-sqrt1-xx2dx", "tags": ["calculus", "integration", "definite-integrals", "closed-form", "hypergeometric-function"], "asker": "Vladimir Reshetnikov", "question_score": 41, "answer_score": 40, "answer_date": "2014-09-04", "type": "integral", "question_markdown": "I want to find a closed form for this integral:\n$$I=\\int_0^1\\frac{\\ln^2x}{\\sqrt{x^2-x+1}}dx\\tag1$$\n_Mathematica_ and _Maple_ cannot evaluate it directly, and I was not able to find it in tables. A numeric approximation for it is\n$$I\\approx2.100290124838430655413586565140170651784798511276914224...\\tag2$$\n(click [here](https://gist.githubusercontent.com/VladimirReshetnikov/88c45337cb2f3129bb77/raw/baa02f2c728abac206a6a88b6b727bd1067b22a5/gistfile1.txt) to see more digits).\n\n_Mathematica_ is able to find a closed form for a parameterized integral in terms of the [Appell hypergeometric function](http://mathworld.wolfram.com/AppellHypergeometricFunction.html):\n$$I(a)=\\int_0^1\\frac{x^a}{\\sqrt{x^2-x+1}}dx\\\\=\\frac1{a+1}F_1\\left(a+1;\\frac{1}{2},\\frac{1}{2};a+2;(-1)^{\\small1/3},-(-1)^{\\small2/3}\\right).\\tag3$$\nI suspect this expression could be rewritten in a simpler form, but I could not find it yet.\n\nIt's easy to see that $I=I''(0),$ but it's unclear how to find a closed-form derivative of the Appell hypergeometric function with respect to its parameters.\n\nCould you help me to find a closed form for $I$?\n\n---\n_Update:_ Numerical calculations suggest that for all complex $z$ with $\\Re(z)>0$ the following functional equation holds:\n$$z\\,I(z-1)-\\!\\left(z+\\tfrac12\\right)\\,I(z)+(z+1)\\,I(z+1)=1.\\tag4$$", "cleo_answer_markdown": "$$I=12\\operatorname{Li}_3\\left(\\frac13\\right)+20\\operatorname{Li}_3\\left(\\frac23\\right)+\\frac{32}3\\operatorname{Li}_2\\left(\\frac23\\right)\\ln3+\\left(4\\ln2+\\frac{10}{3}\\ln3\\right)\\operatorname{Li}_2\\left(\\frac34\\right)\\\\-\\frac{163}6\\zeta(3)+8\\ln^32+\\frac23\\ln^22\\cdot\\ln3-\\frac23\\ln2\\cdot\\ln^23-\\frac{7\\pi^2}3\\ln2+\\frac{11\\pi^2}{9}\\ln3.$$", "n": 12, "slug": "q918821-closed-form-for-large-int-0-1-frac-ln-2x-sqrt-1"}
{"question_id": 1150822, "answer_id": 1150898, "title": "Closed form for $\\int_0^\\infty\\arctan\\Bigl(\\frac{2\\pi}{x-\\ln\\,x+\\ln(\\frac\\pi2)}\\Bigr)\\frac{dx}{x+1}$", "url": "https://math.stackexchange.com/questions/1150822/closed-form-for-int-0-infty-arctan-bigl-frac2-pix-ln-x-ln-frac-pi2", "tags": ["calculus", "integration", "definite-integrals", "logarithms", "closed-form"], "asker": "Laila Podlesny", "question_score": 36, "answer_score": 36, "answer_date": "2015-02-16", "type": "integral", "question_markdown": "I'm trying to find a closed form for this integral:\n$$I=\\int_0^\\infty\\arctan\\left(\\frac{2\\pi}{x-\\ln\\,x+\\ln\\left(\\frac\\pi2\\right)}\\right)\\frac{dx}{x+1}$$\nIts approximate numeric value is\n$$I\\approx3.3805825284453469793953592216276992165696856825906055108192183...$$\n\nAny help is appreciated. Thanks!", "cleo_answer_markdown": "$$I=\\pi\\,\\ln\\left(\\frac{1+\\pi^2+\\ln^2\\left(\\frac\\pi2\\right)-2\\ln\\left(\\frac\\pi2\\right)}{1+\\frac{\\pi^2}4}\\right)$$", "n": 13, "slug": "q1150822-closed-form-for-int-0-infty-arctan-bigl-frac-2-p"}
{"question_id": 1376159, "answer_id": 1384389, "title": "A difficult logarithmic integral ${\\Large\\int}_0^1\\log(x)\\,\\log(2+x)\\,\\log(1+x)\\,\\log\\left(1+x^{-1}\\right)dx$", "url": "https://math.stackexchange.com/questions/1376159/a-difficult-logarithmic-integral-large-int-01-logx-log2x-log1x", "tags": ["integration", "definite-integrals", "logarithms", "closed-form", "polylogarithm"], "asker": "Laila Podlesny", "question_score": 27, "answer_score": 31, "answer_date": "2015-08-04", "type": "integral", "question_markdown": "A friend of mine shared this problem with me. As he was told, this integral can be evaluated in a closed form (the result may involve polylogarithms). Despite all our efforts, so far we have not achieved anything, so I decided to ask for your advice.\n$$\\int_0^1\\log(x)\\,\\log(2+x)\\,\\log(1+x)\\,\\log\\left(1+x^{-1}\\right)dx$$\n\nI found some similar questions here on MSE: \n[(1)](https://math.stackexchange.com/q/316745/76878), \n[(2)](https://math.stackexchange.com/q/465444/76878),\n[(3)](https://math.stackexchange.com/q/503405/76878), [(4)](https://math.stackexchange.com/q/524358/76878),\n[(5)](https://math.stackexchange.com/q/761930/76878),\n[(6)](https://math.stackexchange.com/q/795867/76878),\n[(7)](https://math.stackexchange.com/q/908108/76878), \n[(8)](https://math.stackexchange.com/q/915083/76878), [(9)](https://math.stackexchange.com/q/933977/76878), [(10)](https://math.stackexchange.com/q/972775/76878),\n[(11)](https://math.stackexchange.com/q/1043771/76878), [(12)](https://math.stackexchange.com/q/1046519/76878),\n[(13)](https://math.stackexchange.com/q/1096557/76878), [(14)](https://math.stackexchange.com/q/1341254/76878).", "cleo_answer_markdown": "\\begin{align}\n& \\int_0^1\\ln(2+x)\\,\\ln(1+x)\\,\\ln\\left(1+x^{-1}\\right)\\ln x\\,dx\\\\\n& \\quad=\\frac{71}{36}\\,\\ln^42+2\\ln^32\\cdot\\ln3+4\\ln2\\cdot\\ln^33-7\\ln^22\\cdot\\ln^23-\\frac23\\,\\ln^32-\\frac23\\,\\ln^33-\\ln^22\\cdot\\ln3\\\\\n& \\quad \\quad +6\\ln^22+3\\ln^23-12\\ln2-\\frac{\\pi^4}{216}+\\pi^2\\!\\left(\\frac{49}{36}\\,\\ln^22-2\\ln2\\cdot\\ln3-\\frac{\\ln2}3+\\frac{\\ln3}3-\\frac16\\right)\\\\\n& \\quad \\quad+\\left(6-2\\ln2-2\\ln^22\\right)\\operatorname{Li}_2\\!\\left(\\tfrac13\\right)+(2-12\\ln2)\\left[\\operatorname{Li}_3\\!\\left(\\tfrac13\\right)+\\operatorname{Li}_3\\!\\left(\\tfrac23\\right)\\right]-\\frac23\\,\\operatorname{Li}_4\\!\\left(\\tfrac12\\right)+3\\operatorname{Li}_4\\!\\left(\\tfrac14\\right)\\\\\n& \\quad \\quad +\\left(\\frac54+\\frac{221}{12}\\ln2\\right)\\zeta(3).\n\\end{align}", "n": 14, "slug": "q1376159-a-difficult-logarithmic-integral-large-int-0-1-l"}
{"question_id": 970125, "answer_id": 970624, "title": "Evaluate $\\int_0^1\\frac{\\ln(1-x)}{x}\\text{Li}_3\\left(\\frac{1+x}{2}\\right)dx$ , $\\int_0^1\\frac{\\ln^2(1-x)}{x}\\text{Li}_2\\left(\\frac{1+x}{2} \\right)dx$", "url": "https://math.stackexchange.com/questions/970125/evaluate-int-01-frac-ln1-xx-textli-3-left-frac1x2-rightdx", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "closed-form"], "asker": "xuce1234", "question_score": 17, "answer_score": 30, "answer_date": "2014-10-12", "type": "integral", "question_markdown": "How can we evaluate the following integrals:\n$$\\int_0^1\\frac{\\ln(1-x)}{x}\\text{Li}_3\\left(\\frac{1 + x}{2} \\right)\\,dx\\\\\n.\\\\\n\\int_0^1\\frac{\\ln^2(1-x)}{x}\\text{Li}_2\\left(\\frac{1 + x}{2} \\right)\\,dx$$", "cleo_answer_markdown": "$$\\begin{align*}{\\large\\int}_0^1\\frac{\\ln(1-x)\\,\\operatorname{Li}_3\\left(\\frac{1+x}2\\right)}xdx&=\\frac{29\\,\\zeta(5)}{16}-\\frac{19\\pi^2}{96}\\zeta(3)+\\frac{5\\,\\zeta(3)}{16}\\ln^22+\\frac{\\ln^52}{40}\\\\&-\\frac{5\\pi^2}{72}\\ln^32+\\frac{11\\pi^4}{1440}\\ln2-3\\operatorname{Li}_5\\left(\\tfrac12\\right).\\\\\n\\\\\n{\\large\\int}_0^1\\frac{\\ln^2(1-x)\\,\\operatorname{Li}_2\\left(\\frac{1+x}2\\right)}xdx&=\\frac{81\\,\\zeta(5)}{32}+\\frac{5\\pi^2}{16}\\zeta(3)-\\frac{\\zeta(3)}8\\ln^22+\\frac1{15}\\ln^52\\\\&-\\frac{\\pi^2}{18}\\ln^32-\\frac{\\pi^4}{15}\\ln2+2\\operatorname{Li}_5\\left(\\tfrac12\\right)+2\\operatorname{Li}_4\\left(\\tfrac12\\right)\\ln2.\n\\end{align*}$$", "n": 15, "slug": "q970125-evaluate-int-0-1-frac-ln-1-x-x-text-li-3-left-fr"}
{"question_id": 1372767, "answer_id": 1372876, "title": "Integral $\\int_0^1\\frac{\\log(x)\\log(1+x)}{\\sqrt{1-x}}\\,dx$", "url": "https://math.stackexchange.com/questions/1372767/integral-int-01-frac-logx-log1x-sqrt1-x-dx", "tags": ["calculus", "integration", "definite-integrals", "logarithms", "hypergeometric-function"], "asker": "OlegK", "question_score": 15, "answer_score": 30, "answer_date": "2015-07-24", "type": "integral", "question_markdown": "I'm trying to evaluate this definite integral:\n$$\\int_0^1\\frac{\\log(x) \\log(1+x)}{\\sqrt{1-x}} dx$$\nIt's clear that the result can be expressed in terms of derivatives of a hypergeometric function with respect to its parameters. I obtained the following form:\n$$4 \\left(1 - \\log 2\\right){_2F_1}^{(0,1,0,0)}\\left(1, 0; \\tfrac{3}{2}; -1\\right) - 2 {_2F_1}^{(1,1,0,0)}\\left(1, 0; \\tfrac{3}{2}; -1\\right) - 2{_2F_1}^{(0,1,1,0)}\\left(1, 0; \\tfrac{3}{2}; -1\\right)$$\nIs it possible to expand these derivatives to some explicit form and further simplify this result? Or maybe you could suggest a different way to evaluate this integral that gives a simpler result without going through hypergeometric functions?", "cleo_answer_markdown": "$$\\int_0^1\\frac{\\ln(1+x)\\ln x}{\\sqrt{1-x}}dx=16-8\\ln2+4\\ln^2\\left(1+\\sqrt2\\right)\\\\+\\sqrt2\\left[2\\ln^22+8\\left(\\ln2-1\\right)\\ln\\left(1+\\sqrt2\\right)-\\frac{7\\pi^2}3+16\\operatorname{Li}_2\\!\\left(\\frac1{\\sqrt{2}}\\right)\\right].$$", "n": 16, "slug": "q1372767-integral-int-0-1-frac-log-x-log-1-x-sqrt-1-x-dx"}
{"question_id": 1153708, "answer_id": 1154065, "title": "Closer form for $\\int_0^\\infty\\frac{(\\arctan{x})^2\\log^2({1+x^2})}{x^2}dx$", "url": "https://math.stackexchange.com/questions/1153708/closer-form-for-int-0-infty-frac-arctanx2-log21x2x2dx", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "polylogarithmic-integral"], "asker": "user178256", "question_score": 29, "answer_score": 28, "answer_date": "2015-02-18", "type": "integral", "question_markdown": "I Would like to know the value of this integral. $$\\int_0^\\infty\\frac{(\\arctan{x})^2\\log^2({1+x^2})}{x^2}dx$$\n I think\n$$I=\\frac{a}{b}(\\pi^3\\ln2)+\\frac{c}{d}(\\pi\\ln^32)+\\frac{e}{f}(\\pi\\ln^22)+\\frac{g}{h}(\\pi\\ln2)+\\frac{i}{j}(\\pi^3)+\\frac{k}{m}\\zeta({3})??$$\nWhere a,b,c,d...are integers\nThanks.", "cleo_answer_markdown": "It has a closed form a bit different from what you conjectured:\n$$I=\\frac{4\\pi}3\\ln^32+\\frac{2\\pi^3}3\\ln2+\\frac\\pi2\\zeta(3)$$", "n": 17, "slug": "q1153708-closer-form-for-int-0-infty-frac-arctan-x-2-log"}
{"question_id": 570997, "answer_id": 571075, "title": "Integral $\\int_0^1\\frac{\\ln\\left(x+\\sqrt2\\right)}{\\sqrt{2-x}\\,\\sqrt{1-x}\\,\\sqrt{\\vphantom{1}x}}\\mathrm dx$", "url": "https://math.stackexchange.com/questions/570997/integral-int-01-frac-ln-leftx-sqrt2-right-sqrt2-x-sqrt1-x-sqrt", "tags": ["calculus", "integration", "definite-integrals", "logarithms", "closed-form"], "asker": "Frida Mauer", "question_score": 54, "answer_score": 25, "answer_date": "2013-11-17", "type": "integral", "question_markdown": "Is there a closed form for the integral\n$$\\int_0^1\\frac{\\ln\\left(x+\\sqrt2\\right)}{\\sqrt{2-x}\\,\\sqrt{1-x}\\,\\sqrt{\\vphantom{1}x}}\\mathrm dx.$$\nI do not have a strong reason to be sure it exists, but I would be very interested to see an approach to find one if it does exist.", "cleo_answer_markdown": "$$\\frac{\\pi^{3/2}}{8\\,\\sqrt2}\\cdot\\frac{7\\ln2-\\ln\\left(17-12\\,\\sqrt2\\right)-\\pi}{\\Gamma\\left(\\frac34\\right)^2}$$", "n": 18, "slug": "q570997-integral-int-0-1-frac-ln-left-x-sqrt2-right-sqrt"}
{"question_id": 909228, "answer_id": 909348, "title": "Infinite Series $\\sum_{n=1}^\\infty\\frac{H_n}{n^32^n}$", "url": "https://math.stackexchange.com/questions/909228/infinite-series-sum-n-1-infty-frach-nn32n", "tags": ["real-analysis", "sequences-and-series", "closed-form", "harmonic-numbers", "zeta-functions"], "asker": "OlegK", "question_score": 37, "answer_score": 25, "answer_date": "2014-08-26", "type": "series", "question_markdown": "I'm trying to find a closed form for the following sum\n$$\\sum_{n=1}^\\infty\\frac{H_n}{n^3\\,2^n},$$\nwhere $H_n=\\displaystyle\\sum_{k=1}^n\\frac{1}{k}$ is a harmonic number.\n\nCould you help me with it?", "cleo_answer_markdown": "$$\\sum_{n=1}^\\infty\\frac{H_n}{n^3\\,2^n}=\\frac{\\pi^4}{720}+\\frac{\\ln^42}{24}-\\frac{\\ln2}8\\zeta(3)+\\operatorname{Li}_4\\left(\\frac12\\right).$$", "n": 19, "slug": "q909228-infinite-series-sum-n-1-infty-frac-h-n-n-32-n"}
{"question_id": 1550806, "answer_id": 1550856, "title": "Closed form solution to $\\int_0^1\\arctan^2(x)\\,\\sqrt{x}\\,dx$", "url": "https://math.stackexchange.com/questions/1550806/closed-form-solution-to-int-01-arctan2x-sqrtx-dx", "tags": ["calculus", "integration", "definite-integrals", "closed-form", "trigonometry"], "asker": "Liu Jin Tsai", "question_score": 13, "answer_score": 23, "answer_date": "2015-11-29", "type": "integral", "question_markdown": "I need to compute this integral:\n$$\\int_0^1\\arctan^2(x)\\,\\sqrt{x}\\,dx$$\nI tried integration by parts, and also introducing a parameter $\\arctan(a\\,x)$ and differentiation wrt it, but these approaches did not lead to anything useful. Please help.", "cleo_answer_markdown": "$$\\frac{\\pi^2}{24}-\\frac{2\\pi}3+\\frac1{36\\sqrt{2}}\\left[5\\pi^2+12\\left(4+\\ln\\left(\\frac{1+\\sqrt2}2\\right)\\right)\\left(\\pi-2\\ln\\left(1+\\sqrt2\\right)\\right)-48\\operatorname{Li}_2\\left(\\sqrt2-1\\right)\\right]$$", "n": 20, "slug": "q1550806-closed-form-solution-to-int-0-1-arctan-2-x-sqrt"}
{"question_id": 564816, "answer_id": 570815, "title": "Integral $\\int_0^{\\pi/2}\\arctan^2\\left(\\frac{6\\sin x}{3+\\cos 2x}\\right)\\mathrm dx$", "url": "https://math.stackexchange.com/questions/564816/integral-int-0-pi-2-arctan2-left-frac6-sin-x3-cos-2x-right-mathrm-d", "tags": ["calculus", "integration", "trigonometry", "definite-integrals", "closed-form"], "asker": "Laila Podlesny", "question_score": 38, "answer_score": 22, "answer_date": "2013-11-17", "type": "integral", "question_markdown": "Is it possible to evaluate this integral in a closed form?\n$$I=\\int_0^{\\pi/2}\\arctan^2\\left(\\frac{6\\sin x}{3+\\cos 2x}\\right)\\mathrm dx$$", "cleo_answer_markdown": "$$I=\\frac\\pi2\\left(2\\,\\operatorname{Li}_2\\Big(\\left(1-\\sqrt2\\right)\\left(2-\\sqrt5\\right)\\Big)-2\\,\\operatorname{Li}_2\\Big(\\left(1-\\sqrt2\\right)\\left(\\sqrt5-2\\right)\\Big)\\\\+\\operatorname{Li}_2\\left(3-\\sqrt8\\right)-\\operatorname{Li}_2\\left(\\sqrt8-3\\right)+\\operatorname{Li}_2\\left(9-\\sqrt{80}\\right)-\\operatorname{Li}_2\\left(\\sqrt{80}-9\\right)\\right)$$", "n": 21, "slug": "q564816-integral-int-0-pi-2-arctan-2-left-frac-6-sin-x-3"}
{"question_id": 577849, "answer_id": 578512, "title": "Derivative of the Meijer G-function with respect to one of its parameters", "url": "https://math.stackexchange.com/questions/577849/derivative-of-the-meijer-g-function-with-respect-to-one-of-its-parameters", "tags": ["calculus", "complex-analysis", "derivatives", "special-functions", "closed-form"], "asker": "Vladimir Reshetnikov", "question_score": 17, "answer_score": 19, "answer_date": "2013-11-23", "type": "other", "question_markdown": "Are there any approaches that allow to find a derivative of the [Meijer G-function](http://mathworld.wolfram.com/MeijerG-Function.html) with respect to one of its parameters in a closed form (or at least numerically with a high precision and in reasonable time, with all found digits provably correct)? I am particularly interested in this case:\n$$\\mathcal{D}=\\left.\\partial_\\alpha G_{2,3}^{2,1}\\left(1\\middle|\\begin{array}c1,\\alpha\\\\1,1,0\\end{array}\\right)\\right|_{\\alpha=1}$$", "cleo_answer_markdown": "Yes, it is possible in some cases. For example,\n$$\\begin{align}\\mathcal{D}&={_2F_2}\\left(\\begin{array}c1,1\\\\2,2\\end{array}\\middle|-1\\right)\\\\&=\\gamma-\\operatorname{Ei}(-1),\\end{align}$$\nwhere ${_pF_q}$ is the [generalized hypergeometric function](http://en.wikipedia.org/wiki/Generalized_hypergeometric_function), $\\gamma$ is the [Euler–Mascheroni constant](http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant), and $\\operatorname{Ei}(z)$ is the [exponential integral](http://en.wikipedia.org/wiki/Exponential_integral). In case you need a numeric value,\n$$\\mathcal{D}\\approx0.7965995992970531342836758655425240800732066293468318063837458...$$", "n": 22, "slug": "q577849-derivative-of-the-meijer-g-function-with-respect"}
{"question_id": 557439, "answer_id": 560221, "title": "Integral $\\int_0^\\infty\\frac{\\operatorname{arccot}\\left(\\sqrt{x}-2\\,\\sqrt{x+1}\\right)}{x+1}\\mathrm dx$", "url": "https://math.stackexchange.com/questions/557439/integral-int-0-infty-frac-operatornamearccot-left-sqrtx-2-sqrtx1-r", "tags": ["calculus", "integration", "trigonometry", "improper-integrals", "closed-form"], "asker": "Laila Podlesny", "question_score": 41, "answer_score": 17, "answer_date": "2013-11-09", "type": "integral", "question_markdown": "Is it possible to evaluate this integral in a closed form?\n$$\\int_0^\\infty\\frac{\\operatorname{arccot}\\left(\\sqrt{x}-2\\,\\sqrt{x+1}\\right)}{x+1}\\mathrm dx$$", "cleo_answer_markdown": "$$\\int_0^\\infty\\frac{\\operatorname{arccot}\\left(\\sqrt{x}-2\\,\\sqrt{x+1}\\right)}{x+1}dx=\\pi\\ln\\frac34-\\Im\\operatorname{Li}_2\\frac{3i}4$$", "n": 23, "slug": "q557439-integral-int-0-infty-frac-operatorname-arccot-le"}
{"question_id": 554624, "answer_id": 564369, "title": "Integral $\\int_0^\\infty\\frac{1}{x\\,\\sqrt{2}+\\sqrt{2\\,x^2+1}}\\cdot\\frac{\\log x}{\\sqrt{x^2+1}}\\mathrm dx$", "url": "https://math.stackexchange.com/questions/554624/integral-int-0-infty-frac1x-sqrt2-sqrt2-x21-cdot-frac-log-x", "tags": ["calculus", "integration", "logarithms", "improper-integrals", "closed-form"], "asker": "X.C.", "question_score": 24, "answer_score": 16, "answer_date": "2013-11-12", "type": "integral", "question_markdown": "I need your assistance with evaluating the integral\n$$\\int_0^\\infty\\frac{1}{x\\,\\sqrt{2}+\\sqrt{2\\,x^2+1}}\\cdot\\frac{\\log x}{\\sqrt{x^2+1}}dx$$\n\nI tried manual integration by parts, but it seemed to only complicate the integrand more. I also tried to evaluate it with a CAS, but it was not able to handle it.", "cleo_answer_markdown": "$$\\frac{1+\\ln\\sqrt[4]2}{\\sqrt{2\\,\\pi}}\\Gamma\\left(\\frac34\\right)^2-\\frac{\\sqrt{2\\,\\pi^3}}8\\Gamma\\left(\\frac34\\right)^{-2}+(\\ln2-1)\\,\\sqrt2$$", "n": 24, "slug": "q554624-integral-int-0-infty-frac-1-x-sqrt-2-sqrt-2-x-2"}
{"question_id": 927427, "answer_id": 929215, "title": "Does $\\int_0^1\\frac{\\ln x}{1+x}\\cos^{-1}x\\,\\mathrm dx$ have a closed from?", "url": "https://math.stackexchange.com/questions/927427/does-int-01-frac-ln-x1x-cos-1x-mathrm-dx-have-a-closed-from", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "closed-form"], "asker": "Anastasiya-Romanova 秀", "question_score": 23, "answer_score": 16, "answer_date": "2014-09-12", "type": "integral", "question_markdown": "Does the following integral have a closed-form ?: \n>\\begin{equation}\n\\int_{0}^{1}{\\ln\\left(\\,x\\,\\right) \\over 1 + x}\\,\\arccos\\left(\\,x\\,\\right)\n\\,{\\rm d}x\n\\end{equation}\n\nThis integral has been posted in [Integral and Series](http://integralsandseries.prophpbb.com/topic459.html) a week ago but it remains unsolved. So, I decide to post it here.\n\nCould anyone here please help me to find the closed-form preferably with elementary ways ( high school methods ) ?.\n\nAny help would be greatly appreciated. Thank you.", "cleo_answer_markdown": "$$\\int_0^1\\frac{\\ln x\\cdot\\arccos x}{1+x}\\,dx=2\\,G\\ln2+\\frac\\pi2\\ln^22+\\frac{\\pi^3}{16}-4\\,\\Im\\operatorname{Li}_3(1+i).$$\nYou also can have a different form that does not use complex numbers:\n$$\\int_0^1\\frac{\\ln x\\cdot\\arccos x}{1+x}\\,dx=2\\,G\\ln2+\\frac\\pi4\\ln^22-\\frac{\\pi^3}8+{_4F_3}\\!\\left(\\begin{array}c\\tfrac12,\\tfrac12,1,1\\\\\\tfrac32,\\tfrac32,\\tfrac32\\end{array}\\middle|\\,1\\right)\\!.$$", "n": 25, "slug": "q927427-does-int-0-1-frac-ln-x-1-x-cos-1-x-mathrm-dx-hav"}
{"question_id": 577704, "answer_id": 577766, "title": "Integral $\\int_0^\\infty\\frac{\\ln\\left(1+x+\\sqrt{x^2+2\\,x}\\right)\\,\\ln\\left(1+\\sqrt{x^2+2\\,x+2}\\right)}{x^2+2x+1}dx$", "url": "https://math.stackexchange.com/questions/577704/integral-int-0-infty-frac-ln-left1x-sqrtx22-x-right-ln-left1-sq", "tags": ["calculus", "integration", "logarithms", "improper-integrals", "closed-form"], "asker": "X.C.", "question_score": 25, "answer_score": 15, "answer_date": "2013-11-23", "type": "integral", "question_markdown": "Could you suggest any ideas how to evaluate this integral? Is there a closed-form result?\n$$\\int_0^\\infty\\frac{\\ln\\left(1+x+\\sqrt{x^2+2\\,x}\\right)\\,\\ln\\left(1+\\sqrt{x^2+2\\,x+2}\\right)}{x^2+2x+1}dx$$", "cleo_answer_markdown": "This is a very interesting integral! I'll give you a partial answer:  \nYes, there is a closed form expression for it:\n$$G+\\frac\\pi4+\\frac{\\pi+1}2\\ln2,$$\nwhere $G$ is the [Catalan constant](http://en.wikipedia.org/wiki/Catalan's_constant):\n$$G=-\\int_0^1\\frac{\\ln x}{x^2+1}dx.$$", "n": 26, "slug": "q577704-integral-int-0-infty-frac-ln-left-1-x-sqrt-x-2-2"}
{"question_id": 578605, "answer_id": 579584, "title": "Integral $\\int_0^\\infty\\frac{\\ln\\left(\\sqrt{x+1\\vphantom{x^0}}-1\\right)\\,\\ln\\left(\\sqrt{x^{-1}+1}+1\\right)}{(x+1)^{3/2}}dx$", "url": "https://math.stackexchange.com/questions/578605/integral-int-0-infty-frac-ln-left-sqrtx1-vphantomx0-1-right-ln-lef", "tags": ["calculus", "integration", "logarithms", "improper-integrals", "closed-form"], "asker": "X.C.", "question_score": 19, "answer_score": 15, "answer_date": "2013-11-24", "type": "integral", "question_markdown": "Another integral similar to my [previous question](https://math.stackexchange.com/q/577704/79756):\n$$\\int_0^\\infty\\frac{\\ln\\left(\\sqrt{x+1\\vphantom{x^0}}-1\\right)\\,\\ln\\left(\\sqrt{x^{-1}+1}+1\\right)}{(x+1)^{3/2}}dx$$\nCan someone suggest how to evaluate it? Is there a closed form?", "cleo_answer_markdown": "Yes, there is a closed form:\n$$\\frac{\\pi^2}3-\\ln^22-4\\,G,$$\nwhere $G$ is the [Catalan constant](http://en.wikipedia.org/wiki/Catalan's_constant):\n$$G=-\\int_0^1\\frac{\\ln x}{x^2+1}dx.$$", "n": 27, "slug": "q578605-integral-int-0-infty-frac-ln-left-sqrt-x-1-vphan"}
{"question_id": 702681, "answer_id": 703244, "title": "Integral $\\int_0^1\\frac{\\log(1-x)}{\\sqrt{x-x^3}}dx$", "url": "https://math.stackexchange.com/questions/702681/integral-int-01-frac-log1-x-sqrtx-x3dx", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "closed-form"], "asker": "Marty Colos", "question_score": 19, "answer_score": 15, "answer_date": "2014-03-07", "type": "integral", "question_markdown": "I have a trouble with this integral\n$$I=\\int_0^1\\frac{\\log(1-x)}{\\sqrt{x-x^3}}dx.$$\nCould you suggest how to evaluate it?", "cleo_answer_markdown": "$$I=\\frac{\\Gamma\\left(\\frac14\\right)^2}{4\\,\\sqrt{2\\,\\pi}}\\Big(\\ln2-\\pi\\Big).$$", "n": 28, "slug": "q702681-integral-int-0-1-frac-log-1-x-sqrt-x-x-3-dx"}
{"question_id": 935366, "answer_id": 940809, "title": "Simplification of an expression containing $\\operatorname{Li}_3(x)$ terms", "url": "https://math.stackexchange.com/questions/935366/simplification-of-an-expression-containing-operatornameli-3x-terms", "tags": ["calculus", "logarithms", "special-functions", "polylogarithm"], "asker": "Oksana Gimmel", "question_score": 7, "answer_score": 14, "answer_date": "2014-09-21", "type": "other", "question_markdown": "In my computations I ended up with this result:\n$$\\mathcal{K}=78\\operatorname{Li}_3\\left(\\frac13\\right)+15\\operatorname{Li}_3\\left(\\frac23\\right)-64\\operatorname{Li}_3\\left(\\frac15\\right)-102\n\\operatorname{Li}_3\\left(\\frac25\\right)+126\\operatorname{Li}_3\\left(\\frac35\\right)\\\\+12\\operatorname{Li}_3\\left(\\frac45\\right)-89\\operatorname{Li}_3\\left(\\frac16\\right)-152\\operatorname{Li}_3\\left(\\frac56\\right)+63\\operatorname{Li}_3\\left(\\frac38\\right)+76\\operatorname{Li}_3\\left(\\frac58\\right).$$\nI wonder if it's possible to simplify this expression somehow, e.g. to combine some trilogarithm terms to express them using logarithms, or at least to reduce the number of terms?\n\nI tried to apply identities given at [_MathWorld_](http://mathworld.wolfram.com/Trilogarithm.html) and [_Wolfram Functions_](http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/17/ShowAll.html), but could not make the overall expression simpler. _Mathematica_ could not simplify it either.", "cleo_answer_markdown": "Surprisingly, $\\mathcal K$ can be expressed in elementary terms. Let, $$a = \\ln 2\\\\ b=\\ln 3\\\\ c=\\ln 5$$\nThen,\n$$\\mathcal{K}=\\frac23(878 a^3 - 37 b^3 - 7 c^3) - 2 a^2 (202 b + 133 c) + 4 b^2 (-32 a + 19 c) + 3 c^2(13 a - 21 b) + 278 a b c - \\frac23 \\pi^2 (22 a - 50 b + 25 c) \\approx -7.809651$$", "n": 29, "slug": "q935366-simplification-of-an-expression-containing-opera"}
{"question_id": 967398, "answer_id": 967524, "title": "Extract real and imaginary parts of $\\operatorname{Li}_2\\left(i\\left(2\\pm\\sqrt3\\right)\\right)$", "url": "https://math.stackexchange.com/questions/967398/extract-real-and-imaginary-parts-of-operatornameli-2-lefti-left2-pm-sqrt3", "tags": ["calculus", "complex-analysis", "special-functions", "closed-form", "polylogarithm"], "asker": "OlegK", "question_score": 19, "answer_score": 13, "answer_date": "2014-10-11", "type": "other", "question_markdown": "We know that polylogarithms of complex argument sometimes have simple real and imaginary parts, e.g.\n$$\\operatorname{Re}\\big[\\operatorname{Li}_2\\left(i\\right)\\big]=-\\frac{\\pi^2}{48},\\hspace{1em}\\operatorname{Im}\\big[\\operatorname{Li}_2\\left(i\\right)\\big]={\\bf G},$$\nwhere ${\\bf G}$ is the Catalan constant.\n\nAre there closed forms (free of polylogs and imaginary numbers) for any of the following expressions?\n$$\\operatorname{Re}\\big[\\operatorname{Li}_2\\left(i\\left(2\\pm\\sqrt3\\right)\\right)\\big],\\hspace{1em}\\operatorname{Im}\\big[\\operatorname{Li}_2\\left(i\\left(2\\pm\\sqrt3\\right)\\right)\\big]$$", "cleo_answer_markdown": "$$\\Im\\operatorname{Li}_2\\left(i\\left(2\\pm\\sqrt3\\right)\\right)=\\frac{2\\,{\\bf G}}3-\\frac{\\pi\\,(2\\pm3)}{12}\\ln\\left(2-\\sqrt3\\right).$$\nFor the real part so far I could only establish the following identity:\n$$\\Re\\big[\\operatorname{Li}_2\\left(i\\left(2+\\sqrt3\\right)\\right)+\\operatorname{Li}_2\\left(i\\left(2-\\sqrt3\\right)\\right)\\big]=-\\frac{\\pi^2}{24}-\\frac{\\ln^2\\left(2-\\sqrt3\\right)}2.$$", "n": 30, "slug": "q967398-extract-real-and-imaginary-parts-of-operatorname"}
{"question_id": 930011, "answer_id": 931680, "title": "Closed-forms for several tough integrals", "url": "https://math.stackexchange.com/questions/930011/closed-forms-for-several-tough-integrals", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "closed-form"], "asker": "Tunk-Fey", "question_score": 18, "answer_score": 12, "answer_date": "2014-09-15", "type": "integral", "question_markdown": "These integrals came up in the process of finding solution to [Vladimir Reshetnikov's problem](https://math.stackexchange.com/q/921641/123277). I wonder if there are closed-forms for the following integrals:\n\\begin{array}{1,1}\n&[\\text{1}] &\\quad\\int_0^1\\frac{\\operatorname{Li}_3(ax)}{1+2x}\\ dx\\\\[12pt]\n&[\\text{2}] &\\quad\\int_0^1\\frac{\\operatorname{Li}_2(ax)\\ln x}{1+2x}\\ dx\\\\[12pt]\n&[\\text{3}] &\\quad\\int_0^1\\frac{\\ln(1-ax)\\ln^2 x}{1+2x}\\ dx\n\\end{array}\nI have tried many substitutions, integration by parts, or differentiation under integral sign method, but without success so far. I do **not** need a complete or rigorous answer and your answer can be only *Mathematica*'s or *Maple*'s output since I don't have those software packages in my computer or **links** of related papers. I'd be grateful for any help you are able to provide.", "cleo_answer_markdown": "Here are values of the integral $[3]$ for some specific values of the parameter $a$:\n$$\\int_0^1\\frac{\\ln(1-x)\\ln^2x}{1+2x}dx=2\\operatorname{Li}_4\\left(\\frac12\\right)-\\operatorname{Li}_4\\left(\\frac13\\right)-\\operatorname{Li}_4\\left(\\frac23\\right)-\\frac14\\operatorname{Li}_4\\left(\\frac14\\right)\\\\-\\frac{\\ln^42}{12}-\\frac{\\ln^43}{12}+\\frac16\\ln2\\cdot\\ln^33+\\frac{\\pi^2}6\\left(\\ln2\\cdot\\ln3-\\ln^22-\\operatorname{Li}_2\\left(\\frac13\\right)\\right).\\tag1$$\n\n\n>$$\\int_0^1\\frac{\\ln(1+x)\\ln^2x}{1+2x}dx=3\\operatorname{Li}_4\\left(\\frac12\\right)-\\frac34\\operatorname{Li}_4\\left(\\frac14\\right)\\\\+\\left(7\\zeta(3)-\\operatorname{Li}_3\\left(\\frac14\\right)\\right)\\frac{\\ln2}4-\\frac{3\\pi^4}{160}-\\frac{\\ln^42}{24}-\\frac{\\pi^2}6\\ln^22.\\tag2$$\n\n$$\\int_0^1\\frac{\\ln(1+2x)\\ln^2x}{1+2x}dx=\\operatorname{Li}_4\\left(\\frac12\\right)+\\operatorname{Li}_4\\left(\\frac13\\right)+\\operatorname{Li}_4\\left(\\frac23\\right)-\\frac18\\operatorname{Li}_4\\left(\\frac14\\right)\\\\+\\left(\\operatorname{Li}_3\\left(\\frac13\\right)+\\operatorname{Li}_3\\left(\\frac23\\right)\\right)\\ln3-\\frac{11\\pi^4}{360}-\\frac{\\ln^42}{24}-\\frac{\\ln^43}4\\\\+\\frac{\\pi^2}{12}\\left(\\ln^23-\\ln^22\\right)+\\frac13\\ln2\\cdot\\ln^33.\\tag3$$", "n": 31, "slug": "q930011-closed-forms-for-several-tough-integrals"}
{"question_id": 964438, "answer_id": 964594, "title": "Integral $\\int_0^1\\frac{x^{42}}{\\sqrt{x^4-x^2+1}}\\operatorname d \\!x$", "url": "https://math.stackexchange.com/questions/964438/integral-int-01-fracx42-sqrtx4-x21-operatorname-d-x", "tags": ["calculus", "integration", "definite-integrals", "closed-form"], "asker": "Piotr Shatalin", "question_score": 10, "answer_score": 12, "answer_date": "2014-10-09", "type": "integral", "question_markdown": "Could you please help me with this integral?\n$$\\int_0^1\\frac{x^{42}}{\\sqrt{x^4-x^2+1}} \\operatorname d \\!x$$\n\n___\nUpdate: [user153012](https://math.stackexchange.com/users/153012/user153012) posted a result given by a computer that contains scary Appel function, and [Cleo](https://math.stackexchange.com/users/97378/cleo) gave much simpler closed forms for powers $n=42,\\,43$. I am looking for a way to prove those forms. I also would like to find a more general result that would work for arbitrary integer powers, not just $42$.", "cleo_answer_markdown": "$$\\int_0^1\\frac{x^{42}}{\\sqrt{x^4-x^2+1}}dx=\\frac{250\\,351\\,656\\,060\\,403}{1\\,955\\,894\\,551\\,246\\,350}\\\\-\\frac{25\\,556\\,904\\,389\\,521}{391\\,178\\,910\\,249\\,270}{\\bf K}\\left(\\frac{\\sqrt3}2\\right)+\\frac{29\\,595\\,166\\,842\\,073}{977\\,947\\,275\\,623\\,175}{\\bf E}\\left(\\frac{\\sqrt3}2\\right),$$\n\nwhere ${\\bf K}(x)$ and ${\\bf E}(x)$ are complete [elliptic integrals](http://en.wikipedia.org/wiki/Elliptic_integral) of the $1^{st}$ and $2^{nd}$ kind.\n\n---\nYou might be also interested to know that\n$$\\int_0^1\\frac{x^{43}}{\\sqrt{x^4-x^2+1}}dx=\\frac{10\\,495\\,168\\,793\\,593}{86\\,586\\,540\\,687\\,360}-\\frac{98\\,084\\,055\\,671}{1\\,099\\,511\\,627\\,776}\\ln3.$$", "n": 32, "slug": "q964438-integral-int-0-1-frac-x-42-sqrt-x-4-x-2-1-operat", "sanitized": true}
{"question_id": 572859, "answer_id": 573515, "title": "Closed form for $\\int_0^\\infty\\frac{\\sin x\\,\\cdot\\,\\operatorname{Ci}x-\\cos x\\,\\cdot\\,\\operatorname{Si}x}{\\sqrt{16\\,x^2+1}}dx$", "url": "https://math.stackexchange.com/questions/572859/closed-form-for-int-0-infty-frac-sin-x-cdot-operatornamecix-cos-x-c", "tags": ["calculus", "integration", "definite-integrals", "improper-integrals", "closed-form"], "asker": "Marty Colos", "question_score": 12, "answer_score": 11, "answer_date": "2013-11-19", "type": "integral", "question_markdown": "Is it possible to find a closed form for this integral?\n$$\\mathcal{S}=\\int_0^\\infty\\frac{\\sin x\\cdot\\operatorname{Ci}x-\\cos x\\cdot\\operatorname{Si}x}{\\sqrt{16\\,x^2+1}}dx,$$\nwhere $\\operatorname{Ci}x$ is the [cosine integral](http://mathworld.wolfram.com/CosineIntegral.html) and $\\operatorname{Si}x$ is the [sine integral](http://mathworld.wolfram.com/SineIntegral.html):\n$$\\operatorname{Ci}x=-\\int_x^\\infty\\frac{\\cos t}t dt,\\ \\operatorname{Si}x=\\int_0^x\\frac{\\sin t}t dt.$$\nNumerical integration gives\n$$\\mathcal{S}\\approx0.133456902778362645676629...$$", "cleo_answer_markdown": "Not sure if you consider this a closed form, but the integral $\\mathcal{S}$ can be expressed in terms of the generalized Meijer $G$-function (see formula $(3)$ [here](http://mathworld.wolfram.com/MeijerG-Function.html) for the definition) and [the modified Bessel function of the $2^{nd}$ kind $K_\\nu(x)$](http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html):\n$$\\mathcal{S}=\\frac{G_{2,4}^{4,2}\\left(\\frac18,\\frac12\\middle|\\begin{array}{c}\\frac12,\\frac12\\\\0,0,\\frac12,\\frac12\\\\\\end{array}\\right)}{16\\,\\pi}-\\frac\\pi8K_0\\left(\\frac14\\right).$$", "n": 33, "slug": "q572859-closed-form-for-int-0-infty-frac-sin-x-cdot-oper"}
{"question_id": 581215, "answer_id": 582244, "title": "Integral $\\int_0^\\infty x^2\\,e^{-x^2}\\operatorname{erf}(x)\\,\\log(x)\\,dx$", "url": "https://math.stackexchange.com/questions/581215/integral-int-0-infty-x2-e-x2-operatornameerfx-logx-dx", "tags": ["integration", "definite-integrals", "improper-integrals", "closed-form", "hypergeometric-function"], "asker": "Piotr Shatalin", "question_score": 14, "answer_score": 9, "answer_date": "2013-11-26", "type": "integral", "question_markdown": "I need to evaluate this integral:\n$$I=\\int_0^\\infty x^2\\,e^{-x^2}\\operatorname{erf}(x)\\,\\log(x)\\,dx\\tag1$$\nI tried to do this in _Mathematica_ and it returned a result of the form\n$$I=\\frac{(\\pi+2)\\,(1-\\gamma)}{16\\,\\sqrt\\pi}+\\frac1{2\\,\\sqrt\\pi}\\left.\\frac{d}{d\\xi}\\Bigg({_2F_1}\\left(\\tfrac12,\\xi;\\tfrac32;-1\\right)\\Bigg)\\right|_{\\xi=2}\\tag2$$\nI tried to find a closed form for the derivative using an integral representation of the hypergeometric function, but this way returned me back to my original integral. \n\n> Is it possible to represent $I$ in a closed form that does not contain unevaluated integrals or derivatives?", "cleo_answer_markdown": "$$I=\\frac{2-\\ln2}{16}\\sqrt\\pi-\\frac{\\gamma+\\ln2}{16\\,\\sqrt\\pi}(\\pi+2)+\\frac{G}{4\\,\\sqrt\\pi},$$\nwhere $\\gamma$ is the [Euler-Mascheroni constant](http://mathworld.wolfram.com/Euler-MascheroniConstant.html):\n$$\\gamma=-\\int_0^1\\ln(-\\ln x)\\,dx$$\nand $G$ is the [Catalan constant](http://mathworld.wolfram.com/CatalansConstant.html):\n\n$$G=-\\int_0^1\\frac{\\ln x}{x^2+1}dx$$", "n": 34, "slug": "q581215-integral-int-0-infty-x-2-e-x-2-operatorname-erf"}
{"question_id": 904320, "answer_id": 904652, "title": "Improper integral containing $\\sqrt{\\cos x-\\frac1{\\sqrt 2}}$ in the denominator", "url": "https://math.stackexchange.com/questions/904320/improper-integral-containing-sqrt-cos-x-frac1-sqrt-2-in-the-denominator", "tags": ["integration", "definite-integrals", "improper-integrals"], "asker": "user1001001", "question_score": 9, "answer_score": 8, "answer_date": "2014-08-21", "type": "integral", "question_markdown": "How do I find the value of this integral--\n\n$$I=\\displaystyle\\int_{0}^{\\pi/4} \\frac{\\sec^2 x \\ dx}{\\sqrt {\\cos x-\\dfrac{1}{\\sqrt 2}}}$$\n\nI came across this integral too in physics.", "cleo_answer_markdown": "$$I=\\frac\\pi{\\sqrt[4]2}+\\sqrt{20-14\\sqrt2}\\ K\\!\\left(2\\sqrt2-3\\right)+\\sqrt{2+\\sqrt2}\\ E\\!\\left(2\\sqrt2-3\\right)\\\\-2\\sqrt{2-\\sqrt2}\\ \\Pi\\left(\\sqrt2-1,\\,2\\sqrt2-3\\right)$$\nwhere $K(m), E(m), \\Pi(n,m)$ are the [complete elliptic integrals](http://en.wikipedia.org/wiki/Elliptic_integral) of the first, second and third kind:\n$$K(m)={\\large\\int}_0^{\\pi/2}\\frac{d\\theta}{\\sqrt{1-m\\sin^2\\theta}}$$\n$$E(m)={\\large\\int}_0^{\\pi/2}\\sqrt{1-m\\sin^2\\theta}\\,d\\theta$$\n$$\\Pi(n,m)={\\Large\\int}_0^{\\pi/2}\\frac{d\\theta}{\\left(1-n\\sin^2\\theta\\right)\\sqrt {1-m\\sin^2\\theta}}$$\nNote that this notation uses the parameter $m=k^2$ rather than the modulus $k$, that may differ from a convention used in other places.", "n": 35, "slug": "q904320-improper-integral-containing-sqrt-cos-x-frac1-sq"}
{"question_id": 710175, "answer_id": 711804, "title": "Integral $\\int_0^1\\frac{\\ln x}{x-1}\\ln\\left(1+\\frac1{\\ln^2x}\\right)dx$", "url": "https://math.stackexchange.com/questions/710175/integral-int-01-frac-ln-xx-1-ln-left1-frac1-ln2x-rightdx", "tags": ["calculus", "integration", "definite-integrals", "logarithms", "closed-form"], "asker": "Nik Z.", "question_score": 15, "answer_score": 7, "answer_date": "2014-03-14", "type": "integral", "question_markdown": "Is it possible to evaluate this integral in a closed form?\n$$\nI \\equiv \\int_{0}^{1}{\\ln\\left(x\\right) \\over x - 1}\\,\n\\ln\\left(1 + {1 \\over \\ln^{2}\\left(x\\right)}\\right)\\,{\\rm d}x\n$$\nNumerically,\n$$I\\approx2.18083278090426462584033339029703713513\\dots$$", "cleo_answer_markdown": "$$I=\\left(\\pi-\\frac12\\right)\\ln(2\\pi)-\\frac34-4\\pi^2\\left(\\ln\\Gamma\\left(\\frac1{2\\pi}\\right)+\\ln G\\left(\\frac1{2\\pi}\\right)\\right),$$\nwhere $G(z)$ denotes the [Barnes G-function](http://en.wikipedia.org/wiki/Barnes_G-function).", "n": 36, "slug": "q710175-integral-int-0-1-frac-ln-x-x-1-ln-left-1-frac1-l"}
{"question_id": 576304, "answer_id": 576503, "title": "A closed form for $\\int_0^1{_2F_1}\\left(-\\frac{1}{4},\\frac{5}{4};\\,1;\\,\\frac{x}{2}\\right)^2dx$", "url": "https://math.stackexchange.com/questions/576304/a-closed-form-for-int-01-2f-1-left-frac14-frac54-1-fracx", "tags": ["calculus", "integration", "special-functions", "closed-form", "hypergeometric-function"], "asker": "Piotr Shatalin", "question_score": 25, "answer_score": 6, "answer_date": "2013-11-21", "type": "integral", "question_markdown": "Is it possible to evaluate in a closed form integrals containing a squared hypergeometric function, like in this example?\n$$\\begin{align}S&=\\int_0^1{_2F_1}\\left(-\\frac{1}{4},\\frac{5}{4};\\,1;\\,\\frac{x}{2}\\right)^2dx\\\\\\vphantom{=}\\\\&=\\frac{1}{4\\pi}\\int_0^1\\left(\\sum_{n=0}^\\infty\\frac{4n+1}{8^n}\\cdot\\frac{\\Gamma\\left(2n-\\frac{1}{2}\\right)}{\\Gamma(n+1)^2}\\cdot x^n\\right)^2dx\\end{align}$$\nIt is approximately\n$$S\\approx0.8263551866500213413164525287...$$", "cleo_answer_markdown": "Yes, it is possible in some cases, for example,\n$$S=\\frac{8\\sqrt2+4\\ln\\left(\\sqrt2-1\\right)}{3\\pi}$$", "n": 37, "slug": "q576304-a-closed-form-for-int-0-1-2f-1-left-frac-1-4-fra"}
{"question_id": 704917, "answer_id": 705860, "title": "Integral $\\int_0^\\infty F(x)\\,F\\left(x\\,\\sqrt2\\right)\\frac{e^{-x^2}}{x^2} \\, dx$ involving Dawson's function", "url": "https://math.stackexchange.com/questions/704917/integral-int-0-infty-fx-f-leftx-sqrt2-right-frace-x2x2-dx", "tags": ["calculus", "integration", "definite-integrals", "special-functions", "error-function"], "asker": "TauMu", "question_score": 10, "answer_score": 4, "answer_date": "2014-03-09", "type": "integral", "question_markdown": "I need your help evaluating this integral:\n$$I=\\int_0^\\infty F(x)\\,F\\left(x\\,\\sqrt2\\right)\\frac{e^{-x^2}}{x^2} \\, dx,\\tag1$$\nwhere $F(x)$ represents [Dawson's function/integral](http://mathworld.wolfram.com/DawsonsIntegral.html):\n$$F(x)=e^{-x^2}\\int_0^x e^{y^2} \\, dy = \\frac{\\sqrt{\\pi}}{2} e^{-x^{2}} \\operatorname{erfi}(x).\\tag2$$\n\nDawson's function can also be represented by the infinite integral $$F(x) = \\frac{1}{2} \\int_{0}^{\\infty} e^{-t^{2}/4} \\sin(xt) \\,  dt.$$\n\nSince $F(x)$ behaves like $x$ near $x=0$ and like $\\frac{1}{2x}$ for large values of $x$, we know that integral $(1)$ converges.", "cleo_answer_markdown": "$$I=\\frac{\\pi^{3/2}}8\\left(\\sqrt2-4\\right)+\\frac{3\\,\\pi^{1/2}}2\\arctan\\sqrt2$$", "n": 38, "slug": "q704917-integral-int-0-infty-f-x-f-left-x-sqrt2-right-fr"}
{"question_id": 566513, "answer_id": 567070, "title": "Closed form for $\\int_0^1\\sqrt{\\frac{2-x}{(1-x)\\,x}}\\,\\log\\left(\\frac{(2-x)\\,x}{1-x}\\right)dx$", "url": "https://math.stackexchange.com/questions/566513/closed-form-for-int-01-sqrt-frac2-x1-x-x-log-left-frac2-x-x", "tags": ["calculus", "integration", "definite-integrals", "logarithms", "closed-form"], "asker": "Liu Jin Tsai", "question_score": 15, "answer_score": 3, "answer_date": "2013-11-14", "type": "integral", "question_markdown": "This is somewhat similar to my previous question: https://math.stackexchange.com/q/565668/19661\n\nIs it possible to find a closed form for this integral?\n$$Q=\\int_0^1\\sqrt{\\frac{2-x}{(1-x)\\,x}}\\,\\log\\left(\\frac{(2-x)\\,x}{1-x}\\right)dx$$", "cleo_answer_markdown": "$$Q=\\frac{\\Gamma\\left(\\frac34\\right)^{-2}\\pi^2\\ln2-\\Gamma\\left(\\frac34\\right)^2(4-\\ln4)}{\\sqrt{2\\,\\pi}}$$", "n": 39, "slug": "q566513-closed-form-for-int-0-1-sqrt-frac-2-x-1-x-x-log"}