I've tried using the Jacobian matrix, but this method yielded no significant results. Another way is to turn this integral into the $$ \sum_{n=0}^{\infty}\frac{1}{(n+1)^22^n} $$
However, I do not know how to find the exact sum of this series.
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2see e.g. https://en.wikipedia.org/wiki/Spence%27s_function – user90369 Jun 13 '19 at 10:01
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Your comment has helped me a lot. I'm going to study Spence's duties. – Mathsource Jun 13 '19 at 10:18
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I am glad that I could help you. ;) – user90369 Jun 13 '19 at 10:22
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1Try the change of variable $u=(x+y)/2,v=(y-x)/2$ – FDP Jun 13 '19 at 10:33
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The answer should be $\frac{\pi^2}{6}-\ln^2 2$ – Yuriy S Jun 13 '19 at 11:07
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See this answer for reference. Another form of the integral is $$I=-2 \int_0^1 \frac{\ln(1-\eta) d\eta}{1+\eta}$$ – Yuriy S Jun 13 '19 at 12:26
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If you integrate with respect to one variable, then apply integration by parts you arrive at this: https://math.stackexchange.com/a/3113395/515527 – Zacky Jun 13 '19 at 21:30
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As has been mentioned in the comments, the function to use here is $$\mathrm{Li}_2(x)=\sum_{k\ge1}\frac{x^k}{k^2}$$ so that your sum is given by $S=\frac12\mathrm{Li}_2(\frac12)$. To evaluate this exactly, we use the formula $$\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\frac{\pi^2}{6}-\ln(z)\ln(1-z)$$ and plug in $z=1/2$ to get $$S=\frac{\pi^2}{24}-\frac{\ln^2(2)}{4}.$$
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