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Evaluate $\int_0^{\infty}\frac{\log( x)}{x^2+a^2} \,dx$ using contour integration; $Re (a) > 0$

I found two questions where a > 0 but in my case I have the following condition: Re a > 0 (It seems like $a$ can be complex). What will be the difference in solutions?

Same questions:

Evaluate $\int_0^{\infty} \frac{\log(x)dx}{x^2+a^2}$ using contour integration

Contour integration of $\frac{\log( x)}{x^2+a^2}$

I tried to find residues of $\int_0^{\infty}\frac{\log^2( x)}{x^2+a^2} \,dx$ in z=$\pm$ia, where a-complex

rpr
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  • I think maybe it's still $\frac{\pi \log(a)}{2a}$. In the link https://math.stackexchange.com/questions/1826381/evaluate-int-0-infty-frac-logxdxx2a2-using-contour-integration. Sangchul Lee's answer still holds for any $\textrm{Re} a > 0$. The only thing needed to check is $\log(ia)$ and $\log(-ia)$'s arguments. Maybe another way to see it is that $\int_{0}^{\infty}\frac{\log(x)}{x^2+a^2}\textrm{d}x$ defines a analytic function for $\textrm{Re} a > 0$, so it concides with $\frac{\pi \log(a)}{2a}$ – onRiv May 22 '22 at 15:30
  • @I already solved this task and got $\frac{\pi log(a)}{2a}$ :) – rpr May 22 '22 at 15:40

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