$\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}$ using residues I'm trying to find $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx$ where $\alpha>0$ is real.  My approach was to take an integral along the real line from $1/R$ to $R$, around the circle counterclockwise to $-R$, along the real line to $-1/R$, and then around the circle clockwise to $1/R$.  I have encountered 2 problems with this:


*

*This path encloses one pole, at $z=\alpha i$.  I found the residue at $z=\alpha i$ to be $\frac{\ln(\alpha)+i\pi/2}{2\alpha i}$.  However, this gives me that $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx=\frac{\pi(\ln(\alpha)+i\pi/2)}{2\alpha}$.  Since I have a real function integrated over the real line, there cannot be an imaginary part.  Where did I go wrong?  (Also, doing a few examples, the correct answer seems to be $\frac{\pi\ln(\alpha)}{2\alpha}$, the same as I have but without the imaginary part.)

*At first chose my path so instead of going all the way around the upper semicircle, it only went 3/4 of the way around, as I wanted to avoid anything that might go wrong with the discontinuity of $\log(x)$ at the negative real axis.  When I do this, though, I get a different answer than before (the denominator of the fraction is $\alpha(1-e^{3\pi i/4})$ instead of $2\alpha$.  What am I doing wrong that gives me different answers?
 A: Integrating
$$\frac{\log^2 z}{z^2+\alpha^2}$$
around a keyhole contour with the branch cut of the logarithm on the positive
real axis and the argument between  $0$ and $2\pi$ we get from the two
residues
$$2\pi i \times
\left(\frac{(\log\alpha+i\pi/2)^2}{2\alpha i}
- \frac{(\log\alpha+i3\pi/2)^2}{2\alpha i}\right)
\\ = \pi \frac{-\log\alpha \times 2i\pi + 2 \pi^2  }{\alpha}.$$
On the other hand the non-vanishing integrals along the contour
contribute
$$- 4\pi i \int_0^\infty \frac{\log x}{x^2+\alpha^2} dx
\quad\text{and}\quad
4\pi^2 \int_0^\infty \frac{1}{x^2+\alpha^2} dx.$$
Comparing real and imaginary parts we thus obtain for the first integral
$$\frac{1}{4\pi} 
\times \pi \times \frac{\log\alpha}{\alpha} 2 \pi
\\ = \frac{\log\alpha}{\alpha} \frac{\pi}{2}.$$
We also get for the bonus integral
$$\int_0^\infty \frac{1}{x^2+\alpha^2} dx
= \frac{1}{4\pi^2} \times \pi\times \frac{2 \pi^2}{\alpha}
= \frac{\pi}{2\alpha}.$$
A: You were on the right track.  We have
$$\begin{align}
\int_{-\infty}^0 \frac{\log x}{x^2+\alpha^2}dx+\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx&=2\pi i \left(\frac{\log (\alpha)+i\pi/2}{2i\alpha}\right)\\\\
&=\frac{\pi\log \alpha}{\alpha}+i\frac{\pi^2}{2\alpha}
\end{align} \tag 1$$
We change variables on the first integral on the left-hand side of $(1)$ by letting $x\to -x$.  Now, being careful to evaluate $\log (-x)=\log (x)+i\pi$ reveals
$$2\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx+i\pi\int_0^{\infty}\frac{1}{x^2+\alpha^2}=\frac{\pi\log \alpha}{\alpha}+i\frac{\pi^2}{2\alpha}\tag 2$$
whereby equating real and imaginary parts yields of $(2)$
$$\bbox[5px,border:2px solid #C0A000]{\int_{0}^{\infty} \frac{\log x}{x^2+\alpha^2}dx=\frac{\pi\log \alpha}{2\alpha}}$$
and
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}\frac{1}{x^2+\alpha^2}=\frac{\pi}{2\alpha}}$$
which agree with the answers reported by @MarkoRiedel and obtained with a semi-circular contour rather than a keyhole contour!
