Can you help me to calculate this integral. $$\begin{align}I&=\int_{0}^{1}\frac{1}{x^2+1}\left(\int_{0}^{1}\frac{dy}{1+xy}\right)dx\\ &=\int_{0}^{1}\frac{ln(x+1)}{x(x^2+1)}dx\\ &=\ldots\end{align}$$ Its clear that this integral is integrable. Is there some specific technique to make the calculation? Thank you very much for your help.


closed as off-topic by John B, Claude Leibovici, gebruiker, choco_addicted, Jean-Claude Arbaut Mar 29 '16 at 10:33

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  • $\begingroup$ Let you know that $\int\frac{1}{x^2+1}dx=\arctan(x)+c.$ $\endgroup$ – Theodoros Mpalis Mar 29 '16 at 8:52
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    $\begingroup$ @TheodorosMpalis we all know it. Thank you anyway $\endgroup$ – A s Mar 29 '16 at 9:10
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    $\begingroup$ @RonGordon Note that $$\frac1{1+x^2}\frac1{1+xy}=\frac1{1+x^2}\frac1{1+y^2}-\frac{x}{1+x^2}\frac{y}{1+y^2}+\frac{y}{1+y^2}\frac{y}{1+xy}$$ hence $I=J^2-K^2+L$ with $$J=\int_0^1\frac1{1+x^2}dx=\frac{\pi}4,\quad K=\int_0^1\frac{x}{1+x^2}dx=\frac{\ln2}2,$$ and $$L=\int_0^1\frac{y}{1+y^2}\left(\int_0^1\frac{y}{1+xy}dx\right) dy=\int_0^1\frac{y}{1+y^2}\ln(1+y)dy.$$ Now, $$I=\int_0^1\frac{1}{x(1+x^2)}\ln(1+x)dx,$$ hence $$I+L=\int_0^1\frac1x\ln(1+x)dx=\int_0^1\sum_{n\geqslant1}(-1)^{n-1}\frac{x^{n-1}}ndx=\sum_{n\geqslant1}(-1)^{n-1}\frac1{n^2}=\frac12\sum_{n\geqslant1}\frac1{n^2},$$ ... $\endgroup$ – Did Mar 29 '16 at 14:02
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    $\begingroup$ ... that is, $$I+L=\frac{\pi^2}{12},$$ and finally, $$I=\frac{7\pi^2}{96}-\frac{(\ln2)^2}8,$$ modulo the typos you will soon find in all this. :-) $\endgroup$ – Did Mar 29 '16 at 14:03
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    $\begingroup$ @RonGordon No, no, my fault (I happen to be rather tired at the moment, irl...). $\endgroup$ – Did Mar 29 '16 at 14:14