Logarithmic integral and Euler sums

Evaluate:
$$\int_0^1 \frac{\ln (1+x^2)}{1+x}\, {\rm d}x$$

Solution



First we reduce the integral to the Euler sums:

$$\begin{aligned} I &= \int_0^1 \frac{\log(1+x^2)}{1+x}\, {\rm d}x \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\int_0^1 \frac{x^{2n}}{1+x}\, {\rm d}x \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n}\left[\psi_0(n+1)-\psi_0 \left( n+\frac{1}{2}\right) \right] \\ &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n} \left( \sum_{k=1}^n\frac{1}{k} +2\log(2) -2\sum_{k=1}^n \frac{1}{2k-1}\right) \\ &=\log^2(2)+\frac{1}{2}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sum_{k=1}^n\frac{1}{k} -\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sum_{k=1}^n\frac{1}{2k-1} \end{aligned}$$

Now let us focus our attention on evaluating these two Euler sums. We'll begin from the second one:
Using the fact that: $\displaystyle  \int_0^1 (-s)^{n-1}\, {\rm d}s= \frac{(-1)^{n+1}}{n}$ and $\displaystyle \int_0^1 r^{2k-2}\, {\rm d}r = \frac{1}{2k-1}$ we have that:

$$\begin{aligned} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sum_{k=1}^n \frac{1}{2k-1} &= \sum_{n=1}^\infty \sum_{k=1}^n \int_0^1 (-s)^{n-1} \, {\rm d}s \int_0^1 r^{2k-2}\, {\rm d}r \\ &=\sum_{k=1}^\infty \sum_{n=k}^\infty \int_0^1 (-s)^{n-1} \, {\rm d}s \int_0^1 r^{2k-2}\, {\rm d}r \\ &=\sum_{k=1}^\infty \int_0^1 \frac{(-s)^{k-1}}{1+s}\, {\rm d}s \int_0^1 r^{2k-2} \, {\rm d}r \\ &=\int_0^1\int_0^1 \sum_{k=1}^\infty \frac{(-s r^2)^{k-1}}{1+s}\, {\rm d}s \, {\rm d}r \\ &=\int_0^1 \int_0^1 \frac{1}{(1+s)(1+s r^2)}\, {\rm d}s \, {\rm d}r \\ &=\int_0^1 \frac{1}{1+s}\frac{\arctan \sqrt{s}}{\sqrt{s}}\, {\rm d}s \\ &=2\int_0^1 \frac{\arctan t}{1+t^2}\, {\rm d}t \\ &=\frac{\pi^2}{16} \end{aligned}$$

In a similar way the other sum is evaluated easily:

$$\begin{aligned} \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}k \sum_{n=1}^k \dfrac1n & = \sum_{k=1}^{\infty} \sum_{n=1}^k \int_0^1 (-x)^{k-1}\, {\rm d}x \int_0^1 y^{n-1} \, {\rm d}y\\ & = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} \int_0^1 (-x)^{k-1}\, {\rm d}x \int_0^1 y^{n-1} \, {\rm d}y\\ & = \sum_{n=1}^{\infty} \int_0^1 \dfrac{(-x)^{n-1}}{1+x}dx \int_0^1 y^{n-1} \, {\rm d}y\\ & = \int_0^1 \int_0^1\sum_{n=1}^{\infty} \dfrac{(-xy)^{n-1}}{1+x}\, {\rm d}x \, {\rm d}y\\ & = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}\, {\rm d}x \, {\rm d}y\\ & = \int_0^1 \int_0^1\dfrac1{(1+x)(1+xy)}\, {\rm d}y \, {\rm d}x\\ & = \int_0^1 \dfrac{\log(1+x)}{x(1+x)} \, {\rm d}x\\ & = \int_0^1 \dfrac{\log(1+x)}{x} \, {\rm d}x - \int_0^1 \dfrac{\log(1+x)}{(1+x)} \, {\rm d}x\\ & = \dfrac{\zeta(2)}2 - \dfrac{\log^2 2}2 \end{aligned}$$

Combining all the above we get that:

$$\int_0^1 \frac{\ln (1+x^2)}{1+x}\, {\rm d}x = \frac{3}{4}\log^2 2 - \frac{\pi^2}{48}$$

ending the evaluation of the integral.