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I'm trying to evaluate this definite integral: $$\int_0^1\frac{\log(x) \log(1+x)}{\sqrt{1-x}} dx$$ It'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: $$4 \left(1 - \log 2\right){_2F_1}^{(0,1,0,0)}\left(1, 0; \tfrac{3}{2}; -1\right) - 2 {_2F_1}^{(1,1,0,0)}\left(1, 0; \tfrac{3}{2}; -1\right) - 2{_2F_1}^{(0,1,1,0)}\left(1, 0; \tfrac{3}{2}; -1\right)$$ Is 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?

OlegK
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  • To many terms (log and di-log) to reasonably be done by hand, you'll have to settle for symbolic systems. – Zach466920 Jul 24 '15 at 18:36
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    @Zach466920 Actually, it's been my consistent experience with integrals like these that quite the opposite is the case. Wolfram Alpha in particualr can be quite bad at recognizing when massive cancellation is possible among even just a moderate handful of polylog terms. The most reliable way of obtaining the most compact final values tends to be by hand. – David H Jul 24 '15 at 19:30
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    First, let $~t=\sqrt{1-x}~$ or $~x=1-t^2,~$ then write $~\ln(2+t^2)=\ln\bigg[2~\bigg(1+\dfrac{t^2}2\bigg)\bigg],~$ and use the properties of the natural logarithm to break up the initial integral into a sum of two other ones. The smaller of these two is trivial; simply expand the integrand into its well-known Mercator series, and reverse the order of summation and integration. The larger one can be evaluated by Mathematica. – Lucian Jul 24 '15 at 19:57

3 Answers3

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$$\int_0^1\frac{\ln(1+x)\ln x}{\sqrt{1-x}}dx=16-8\ln2+4\ln^2\left(1+\sqrt2\right)\\+\sqrt2\left[2\ln^22+8\left(\ln2-1\right)\ln\left(1+\sqrt2\right)-\frac{7\pi^2}3+16\operatorname{Li}_2\!\left(\frac1{\sqrt{2}}\right)\right].$$

Cleo
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The integral may readily be decomposed into a sum of integrals of products of log-linear terms:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln{\left(x\right)}\ln{\left(1+x\right)}}{\sqrt{1-x}}\,\mathrm{d}x\\ &=2\int_{0}^{1}\ln{\left(1-y^2\right)}\ln{\left(2-y^2\right)}\,\mathrm{d}y;~~~\small{\left[\sqrt{1-x}=y\right]}\\ &=2\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(\sqrt{2}-y\right)}\,\mathrm{d}y\\ &~~~~~+2\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(\sqrt{2}+y\right)}\,\mathrm{d}y\\ &~~~~~+2\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(\sqrt{2}-y\right)}\,\mathrm{d}y\\ &~~~~~+2\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(\sqrt{2}+y\right)}\,\mathrm{d}y.\\ \end{align}$$

Each of these four integrals can be resolved in terms of dilogarithms in a systematic manner, for instance by using the general closed forms for two integrals I derive below. Since a final result has already been provided in another response, I leave the plugging-and-chugging step as an exercise to the fearless reader.

Suppose $0<a\land0<a+b$. Then we find:

$$\begin{align} J{(a,b)} &=\int_{0}^{1}\ln{\left(1-y\right)}\ln{\left(a+by\right)}\,\mathrm{d}y\\ &=\int_{0}^{1}\ln{\left(w\right)}\ln{\left(a+b-bw\right)}\,\mathrm{d}w;~~~\small{\left[1-y=w\right]}\\ &=\small{-\int_{0}^{1}\frac{1}{w}\left[\frac{\left(a+b\right)\ln{\left(a+b\right)}}{b}-\frac{bw+\left(a+b-bw\right)\ln{\left(a+b-bw\right)}}{b}\right]\,\mathrm{d}w}\\ &=\int_{0}^{1}\frac{bw-bw\ln{\left(a+b\right)}+\left(a+b-bw\right)\ln{\left(1-\frac{b}{a+b}w\right)}}{bw}\,\mathrm{d}w\\ &=1-\ln{\left(a+b\right)}+\int_{0}^{1}\frac{\left(a+b-bw\right)\ln{\left(1-\frac{b}{a+b}w\right)}}{bw}\,\mathrm{d}w\\ &=1-\ln{\left(a+b\right)}+\int_{0}^{1}\frac{\left(1-cw\right)\ln{\left(1-cw\right)}}{cw}\,\mathrm{d}w;~~~\small{\left[c:=\frac{b}{a+b}\right]}\\ &=\small{1-\ln{\left(a+b\right)}-\int_{0}^{1}\ln{\left(1-cw\right)}\,\mathrm{d}w+\frac{1}{c}\int_{0}^{1}\frac{\ln{\left(1-cw\right)}}{w}\,\mathrm{d}w}\\ &=1-\ln{\left(a+b\right)}+1+\frac{1-c}{c}\ln{\left(1-c\right)}-\frac{1}{c}\,\operatorname{Li}_{2}{\left(c\right)}\\ &=2-\ln{\left(a+b\right)}+\frac{1-c}{c}\ln{\left(1-c\right)}-\frac{1}{c}\,\operatorname{Li}_{2}{\left(c\right)}\\ &=2+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}-\left(a+b\right)\operatorname{Li}_{2}{\left(\frac{b}{a+b}\right)}}{b}\\ \end{align}$$

Suppose $0<a\land0<a+b\land0<a-b$. Then we find:

$$\begin{align} K{(a,b)} &=\int_{0}^{1}\ln{\left(1+y\right)}\ln{\left(a+by\right)}\,\mathrm{d}y\\ &=\small{\ln{(2)}\ln{\left(a+b\right)}-\int_{0}^{1}y\left[\frac{b\ln{\left(1+y\right)}}{a+by}+\frac{\ln{\left(a+by\right)}}{1+y}\right]\,\mathrm{d}y}\\ &=\ln{(2)}\ln{\left(a+b\right)}\\ &~~~~~\small{-\int_{0}^{1}\left[\ln{\left(1+y\right)}-\frac{a\ln{\left(1+y\right)}}{a+by}+\ln{\left(a+by\right)}-\frac{\ln{\left(a+by\right)}}{1+y}\right]\,\mathrm{d}y}\\ &=\small{\ln{(2)}\ln{\left(a+b\right)}-\left(2\ln{(2)}-1\right)+\frac{b+a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~\small{+\int_{0}^{1}\frac{a\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y+\int_{0}^{1}\frac{\ln{\left(a+by\right)}}{1+y}\,\mathrm{d}y}\\ &=\small{2-2\ln{(2)}+\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~\small{+\int_{0}^{1}\frac{a\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y+\ln{(2)}\ln{\left(a+b\right)}-\int_{0}^{1}\frac{b\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y}\\ &=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~\small{+\left(a-b\right)\int_{0}^{1}\frac{\ln{\left(1+y\right)}}{a+by}\,\mathrm{d}y}\\ &=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~\small{+\left(a-b\right)\int_{1}^{2}\frac{\ln{\left(w\right)}}{a-b+bw}\,\mathrm{d}w};~~~\small{\left[1+y=w\right]}\\ &=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~\small{+\int_{0}^{2}\frac{\ln{\left(w\right)}}{1+\frac{b}{a-b}w}\,\mathrm{d}w-\int_{0}^{1}\frac{\ln{\left(w\right)}}{1+\frac{b}{a-b}w}\,\mathrm{d}w}\\ &=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~\small{+2\int_{0}^{1}\frac{\ln{\left(2u\right)}}{1+\frac{2b}{a-b}u}\,\mathrm{d}u-\frac{a-b}{b}\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}};~~~\small{\left[w=2u\right]}\\ &=\small{2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}}{b}}\\ &~~~~~+2\ln{(2)}\int_{0}^{1}\frac{\mathrm{d}u}{1+\frac{2b}{a-b}u}+2\int_{0}^{1}\frac{\ln{\left(u\right)}}{1+\frac{2b}{a-b}u}\,\mathrm{d}u\\ &~~~~~-\frac{a-b}{b}\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}\\ &=2-2\ln{(2)}+2\ln{(2)}\ln{\left(a+b\right)}\\ &~~~~~+\frac{a\ln{\left(a\right)}-\left(a+b\right)\ln{\left(a+b\right)}+\left(a-b\right)\ln{(2)}\ln{\left(\frac{a+b}{a-b}\right)}}{b}\\ &~~~~~+\frac{a-b}{b}\left[\operatorname{Li}_{2}{\left(-\frac{2b}{a-b}\right)}-\operatorname{Li}_{2}{\left(-\frac{b}{a-b}\right)}\right].\\ \end{align}$$

David H
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  • Very nice! How do we get only one dilog term at the end? – Vladimir Reshetnikov Jul 24 '15 at 22:53
  • @VladimirReshetnikov Note that the integrals we need to calculate correspond to plugging in the values $a=\sqrt{2}$ and $b=\pm1$ into the general expressions I derived. When you go to compute the arguments of all the dilog terms required, you'll find that they're all expressible in terms of the silver ratio or its negative/reciprocal/etc., so it's not surprising most of the dilog terms can be eliminated. Also, see my response here. – David H Jul 24 '15 at 23:20
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{1} {\ln\pars{x}\ln\pars{1 + x} \over \root{1 - x}}\,\dd x}} \,\,\,\stackrel{x\ =\ 1 - t^{2}}{=}\,\,\, \int_{-1}^{1}\ln\pars{1 - t^{2}}\ln\pars{2 - t^{2}}\,\dd t \\[5mm] = &\ 2\int_{-1}^{1}\ln\pars{1 - t}\ln\pars{\root{2} - t}\,\dd t + 2\int_{-1}^{1}\ln\pars{1 - t}\ln\pars{\root{2} + t}\,\dd t \\[1cm] = &\ \bracks{\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t + \int_{-1}^{1}\ln^{2}\pars{\root{2} - t}\,\dd t - \int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} - t}\,\dd t} \\[5mm] + &\ \bracks{\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t + \int_{-1}^{1}\ln^{2}\pars{\root{2} + t}\,\dd t - \int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} + t}\,\dd t} \\[1cm] = &\ 2\int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t + 2\int_{-1}^{1}\ln^{2}\pars{\root{2} - t}\,\dd t \\[5mm] - &\ \int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} - t}\,\dd t - \int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} + t}\,\dd t \label{1}\tag{1} \end{align} The first and the second integral, in \eqref{1}, are quite trivial: \begin{align} \int_{-1}^{1}\ln^{2}\pars{1 - t}\,\dd t & = 4 - 4\ln\pars{2} + 2\ln^{2}\pars{2} \\[5mm] \int_{-1}^{1}\ln^{2}\pars{\root{2} - t}\,\dd t & = 4 - 4\root{2}\ln\pars{1 + \root{2}} + 2\ln^{2}\pars{1 + \root{2}} \end{align} With the change of variables $\ds{{1 - t \over \root{2} \pm t} = x}$, the third and fourth integral, in \eqref{1}, can be rewritten as \begin{align} \int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} - t}\,\dd t & = \pars{\root{2} - 1}\int_{0}^{2\root{2} - 2} {\ln^{2}\pars{x} \over \pars{1 - x}^{2}}\,\dd x \\[5mm] \int_{-1}^{1}\ln^{2}\pars{1 - t \over \root{2} + t}\,\dd t & = \pars{\root{2} + 1}\int_{0}^{2\root{2} + 2} {\ln^{2}\pars{x} \over \pars{1 + x}^{2}}\,\dd x \end{align} Both integrals are straightforward evaluated by successive integration by parts.

Felix Marin
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