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I have to prove that $$\int_{0}^{\infty} \frac{\ln(x)}{(x+a)^2+b^2}=\frac{\ln(\sqrt{a^2+b^2})}{b}\arctan(\frac{b}{a})$$ for $a,b \in \mathbb{R}$ using complex analysis.

I have to use the residue theorem and the Jordan's lemma. I know there is an answer here: Contour integral for finding $\int_0^\infty \frac{\ln x}{(x+a)^2+b^2} \, dx$ But in this answer don't use the form of Jordan's lemma, that is $\lim_{R \rightarrow\infty} \int_\gamma f(z)e^{i \lambda z}dz=0$ where $\lambda > 0$ and $\gamma$ is a family of circular arcs of radii $R$ and $Imz \geq a$.
I've tried using the function $$g(z)=\frac{Ln(z)}{(z+a)^2+b^2}$$ and I found that the poles are $-a+ib$ and $-a-ib$. But from here I don't know how to continue to know which branch of the logarithm I should eliminate.
I would appreciate if someone helps me

GHR01
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  • I know you are being asked to use residue theorem, which is why this is a comment and not an answer, but you can do this integral without complex analysis. Simply use the substitution $xy = a^2+b^2$ – Ninad Munshi Jan 19 '21 at 05:50
  • Have you tried differentiating both sides? – Some Guy Jan 19 '21 at 05:51
  • If both derivatives are equal, that means that the original expressions are equal, or only differ by a constant – Some Guy Jan 19 '21 at 05:52
  • @NinadMunshi I have to solve this using the results that I've mentioned – GHR01 Jan 19 '21 at 05:54
  • Yes, it could be a solution, but I have to solve this using the Residue Theorem and the Jordan's lemma@SomeGuy – GHR01 Jan 19 '21 at 05:54
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    Instead of adopting a keyhole contour as your integration path, choose a semicircle on the upper half of the complex plane and remember that $log(z)$ for $z<0$ is equal to $log|-z|+πi$. Then, assuming that $b≠0$ you just need to compute the residue of $-a+bi$. – Teruo Jan 22 '21 at 02:27

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