<s>[INST] Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and  branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
 [/INST] Setting $a=1$ in
$$ \int_0^\infty\frac{\operatorname{Li}_a(-x)}{1+x^2}\mathrm{d}x=-2^{-a-1}\pi\, \eta(a)-a\beta(a+1)$$
we have
$$\int_0^\infty\frac{\operatorname{Li}_1(-x)}{1+x^2}\mathrm{d}x=-\int_0^\infty\frac{\ln(1+x)}{1+x^2}\mathrm{d}x=-\frac14\ln(2)-\beta(2)=-\frac14\ln(2)-G$$

Other results:
\begin{gather*}
 \int_0^\infty\frac{\operatorname{Li}_2(-x)}{1+x^2}\mathrm{d}x=-\frac{\pi^3}{96}-2\beta(3);\\
 \int_0^\infty\frac{\operatorname{Li}_3(-x)}{1+x^2}\mathrm{d}x=-\frac{3\pi}{64}\zeta(3)-3\beta(4);\\
 \int_0^\infty\frac{\operatorname{Li}_4(-x)}{1+x^2}\mathrm{d}x=-\frac{7\pi^5}{23040}-4\beta(5);\\
 \int_0^\infty\frac{\operatorname{Li}_5(-x)}{1+x^2}\mathrm{d}x=-\frac{15\pi}{1024}\zeta(5)-5\beta(6).
 \end{gather*}

Proof: By writing the integral representation of $\operatorname{Li}_a(-x)$, we have
\begin{gather*}
 \int_0^\infty\frac{\operatorname{Li}_a(-x)}{1+x^2}\mathrm{d}x=\int_0^\infty\frac{1}{1+x^2}\left(\frac{(-1)^a}{(a-1)!}\int_0^1\frac{x\ln^{a-1}(y)}{1+xy}\mathrm{d}y\right)\mathrm{d}x\\
 \{\text{change the order of the integration}\}\\
 =\frac{(-1)^a}{(a-1)!}\int_0^1\ln^{a-1}(y)\left(\int_0^\infty\frac{x}{(1+x^2)(1+xy)}\mathrm{d}x\right)\mathrm{d}y\\
 \{\text{compute the inner integral by partial fraction decomposition}\}\\
 =\frac{(-1)^a}{(a-1)!}\int_0^1\ln^{a-1}(y)\left(\frac{\pi}{2}\frac{y}{1+y^2}-\frac{\ln(y)}{1+y^2}\right)\mathrm{d}y\\
 =\frac{(-1)^a\pi}{2(a-1)!}\underbrace{\int_0^1\frac{y\ln^{a-1}(y)}{1+y^2}\mathrm{d}y}_{y=\sqrt{x}}-\frac{(-1)^a}{(a-1)!}\int_0^1\frac{\ln^a(y)}{1+y^2}\mathrm{d}y\\
 =\frac{(-1)^a\pi}{2^{a+1}(a-1)!}\int_0^1\frac{\ln^{a-1}(x)}{1+x}\mathrm{d}x-\frac{(-1)^a}{(a-1)!}\int_0^1\frac{\ln^a(y)}{1+y^2}\mathrm{d}y\\
=-2^{-a-1}\pi\, \eta(a)-a\beta(a).
 \end{gather*}
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