How to calculate $$\int_{0}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx $$ ? How to choose the branch cut and proceed?
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As an option you may take the initial integrand, but make a cut between logarithm branch points: $x=i$ and $x=-i$, and integrate, for example, in the upper half-plane, bypassing the cut. It is more convenient in this case to evaluate $\int_{0}^{\infty }\frac{ln(1+x^{2})}{(a^2+x^{2})} dx$ - just to push apart branche points and poles; then $a\to1$. This is justified, because the integrand does not have any peculiarity at $a=1$ – Svyatoslav Feb 19 '21 at 19:30
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Sorry, corrected some typo. First option: $\int_{0}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx=\frac{1}{2}\int_{-\infty}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx =\Re\frac{1}{2}\int_{-\infty}^{\infty }\frac{ln(i+x)}{(1+x^{2})} dx $. Then close the contour in the upper half-plane, where there are no branch points (ln is single-valued). – Svyatoslav Feb 19 '21 at 20:12
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First, as $\frac{ln(1+x^{2})}{1+x^{2}}$ is an even function, $\int_{0}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx = \frac{1}{2} \int_{- \infty}^{\infty }\frac{ln(1+x^{2})}{(1+x^{2})} dx$.
Now, there's a general theorem which says that if $lim_{z \to \infty}zf(z)=0$ then \begin{equation} \int_{- \infty}^{\infty }f(x) dx = 2 \pi i \sum_{\substack{a \in Sing(f) \\ a \in \mathbb{H}}}Res(f,a) \ + \ \pi i \sum_{\substack{a \in Sing(f) \\ Im(a) = 0}}Res(f,a) \end{equation} where $\mathbb{H}= \{\,z \in \mathbb{C}: Im(z) > 0\,\} $.
You know that $i$ and $-i$ are the only singularities of your function. As $-i \not\in \mathbb{H}$ your problem is reduced to find $Res(f,i)$.
François Diderot
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