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