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I know that $\int_0^1 \frac{\ln (x+1)}{x^2+1}dx$ is a well-known integral and there are many ways of computing it, including substitutions $x=\tan t$, $x = \frac{1-t}{1+t}$ or differentiating the parametrized integral $I(t) = \int_0^1 \frac{\ln(tx+1)}{x^2+1}dx$. I am interested in whether my solution attempt can be finalized. $$\int_0^1\frac{\ln(x+1)}{x^2+1}dx = \int_0^1\ \sum_{i=0}^{\infty}\frac{(-1)^ix^{i+1}}{i+1}\sum_{j=0}^{\infty}(-x^2)^jdx$$ Here I just rewrote $\ln(x+1)$ and $\frac{1}{x^2+1}$ to their Taylor series, which obviously converge to the needed functions since $x\in[0; 1]$. Now I withdraw everything that does not contain $x$ behind the integral sign and compute the simple $\int_0^1 x^n = \frac{1}{n+1}$ integral: $$=\sum_{i=0}^\infty\sum_{j=0}^\infty \frac{(-1)^{i+j}}{i+1}\int_0^1x^{2j+i+1}dx = \sum_{i=0}^\infty\sum_{j=0}^{\infty}\frac{(-1)^{i+j}}{(i+1)(2j+i+2)}$$ I can now part the fraction in two, since $\frac{1}{(i+1)(2j+i+2)}=\frac{1}{2j+1}(\frac{1}{i+1}-\frac{1}{2j+i+2})$, so I have: $$=\sum_{i=0}^\infty\frac{(-1)^i}{i+1}\sum_{j=0}^\infty\frac{(-1)^j}{2j+1} - \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{(-1)^{i+j}}{(2j+1)(2j+i+2)}$$ And by writing Taylor series for $\ln2$ and $\arctan1=\frac{\pi}{4}$ I get: $$=\ln 2\cdot\frac{\pi}{4} - \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{(-1)^{i+j}}{(2j+1)(2j+i+2)}$$ That's where the problem starts. If we knew the value of our integral, which is $\frac{\pi}{8}\ln2$, we would automatically deduce that: $$\sum_{i=0}^\infty\sum_{j=0}^\infty\frac{(-1)^{i+j}}{(i+1)(2j+i+2)} = \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{(-1)^{i+j}}{(2j+1)(2j+i+2)} \textbf{(*)}$$ And vice versa, if we could prove the above equality independently of $\int_0^1\frac{\ln(x+1)}{x^2+1}=\frac{\pi}{8}\ln 2$, we would automatically conclude that $$I=\frac{\pi}{4}\ln 2 - I,$$ where I is equally our integral and both of the sums (*). And we're done as then $I=\frac{\pi}{8}\ln 2$.

So any help in proving (*) would be appreciated, as well as showing other ways to use Taylor series in this problem.

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    $\begingroup$ You have to be careful with double sums that don't converge absolutely. Does $\sum_{i,j=0}^\infty$ mean $\sum_{i=0}^\infty \sum_{j=0}^\infty$ or $\sum_{j=0}^\infty \sum_{i=0}^\infty$? They might not be the same. $\endgroup$ Apr 17, 2020 at 14:54
  • $\begingroup$ @RobertIsrael It's $\sum_{i=0}^\infty\sum_{j=0}^\infty$ and the order of sums is never changed. I will edit the question. $\endgroup$
    – dnes
    Apr 17, 2020 at 15:13

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