How can I prove $\int^{\infty}_{0} \frac{\ln(x)}{(x+a)^2+b^2} dx = \frac{\ln(\sqrt{a^2+b^2})}{b} \arctan(\frac{b}{a})$

Question

I consider the following:

$$f(z) = \frac{\operatorname{Ln} (z)}{(z+a)^2+b^2}$$

And the contour:

Then, $f(z)$ has a simple pole at $z= -a+ib$

But after developing , I didn't get that result. How could I prove it?

Answer 1

You made the mistake of choosing a contour that doesn't end up with your choice of integral. one alternative is choosing the standard keyhole contour with a branch cut from $0 \to \infty$, and restrict $\arg z\in(0,2\pi]$. If $a,b\in\mathfrak{R^{+}}$ now, instead of choosing $\frac{\log(z)}{(z+a)^2+b^2}$, you choose $\frac{\log^{2}(z)}{(z+a)^2+b^2}$, and it will be clear very soon why.

Now by Residue theorem, $$\oint_{C}\frac{\log^{2}(z)}{(z+a)^2+b^2}dz={-4\pi \over b}\left(\ln\left(\sqrt{a^2+b^2}\right)\cdot i\arctan({b \over a})-\pi \arctan({b \over a})\right)$$

When you open $\log(z)=\log|z|+i \arg(z)$, be careful of the fact that $\arg z\in(0,2\pi]$ so if $\arctan({b \over a})=\theta$, then the argument at residues becomes $\pi \pm \theta$ because $a$ and $b$ are both positive and the residues lie in the second and first quadrant. Then we have

$$\implies {-4\pi \over b}\left(\ln\left(\sqrt{a^2+b^2}\right)\cdot i \arctan({b \over a})-\pi \arctan({b \over a})\right)$$$$=\lim_{R \rightarrow \infty,\epsilon\rightarrow0}\int_{\epsilon}^{R}\frac{\log^{2}(z)}{(z+a)^2+b^2}+\int_{R}^{\epsilon}\frac{\left(\log(z)+2\pi i\right)^{2}}{(z+a)^2+b^2}dz$$ and the integrals over outer and inner circles of radius R and $\epsilon$ collapse in the limit. Equate the imaginary and real parts to get $\int_{\mathfrak{R^{+}}}\frac{\log(z)dz}{(z+a)^2+b^2}=\frac{\ln\left(\sqrt{a^2+b^2}\right)}{b}\arctan({b \over a})$ and $\int_{\mathfrak{R^{+}}}\frac{dx}{(x+a)^2+b^2}=\frac{\arctan({\frac{b}{a}})}{b}$