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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.

dnes
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    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. – Robert Israel Apr 17 '20 at 14:54
  • @RobertIsrael It's $\sum_{i=0}^\infty\sum_{j=0}^\infty$ and the order of sums is never changed. I will edit the question. – dnes Apr 17 '20 at 15:13

1 Answers1

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Here's one possible "other way". Instead of trying to multiply the expansions of $\log(1+x)$ and $\dfrac1{1+x}$, one can use the identities in this answer (both derived from Taylor series), plus the multiplication formula for the harmonic numbers, to draw out a related integral.

$$\begin{align*} I &= \int_0^1 \frac{\log(1+x)}{1+x^2} \, dx \\ &= \sum_{n\ge1} \frac{(-1)^{n-1}}{2n-1} \left(H_{2n-1} - H_{n-\frac12}\right) \\ &= \Im \left\{\sum_{n\ge1} \frac{i^n}n H_n\right\} - \sum_{n\ge1} \frac{(-1)^{n-1}}{2n-1} \left(2H_{2n}-H_n-2\log2\right) \\ &= \frac\pi2\log2+\log2 -\frac\pi2- \Im\left\{\sum_{n\ge1} \frac{i^n}n H_n\right\} + \sum_{n\ge1} \frac{(-1)^{n-1}}{2n-1} H_n \\ &= \frac{5\pi}8\log2+\log2-\frac\pi2 - G + \sum_{n\ge1} \frac{(-1)^{n-1}}{2n-1} H_n \\ &= \frac{5\pi}8\log2 +\log2-\frac\pi2 - G - \int_0^1 \frac{\log\left(1+t^2\right)}{t^2\left(1+t^2\right)} \, dt\\ &= \frac{5\pi}8\log2 - G - \underbrace{\int_0^1 \frac{\log\left(1+t^2\right)}{1+t^2} \, dt}_J \end{align*}$$

Upon substituting $t=\tan u$, we can compute $J$ by utilizing a Fourier series (also derived via Taylor):

$$\begin{align*} J &= 2\int_0^\tfrac\pi4\log(\sec u)\,du \\ &= \frac\pi2\log2 + 2 \sum_{k\ge1}\frac{(-1)^k}k \int_0^\tfrac\pi4 \cos(2ku) \, du \\ &= \frac\pi2\log2 + \sum_{k\ge1} \frac{(-1)^k}{k^2} \sin\frac{k\pi}2 \\ &= \frac\pi2\log2 + \Im \sum_{k\ge1} \frac{e^{-ik\pi/2}}{k^2} \\ &= \frac\pi2\log2 + \Im \operatorname{Li}_2(-i) = \frac\pi2\log2-G \end{align*}$$

Hence $I=\dfrac\pi8\log2$.

user170231
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