integrate the following function: $\int_{0}^{\infty} \frac{\log{x}}{x^{2}+a^{2}} \ dx$ How do I integrate the following function: $$\int_{0}^{\infty} \frac{\log{x}}{x^{2}+a^{2}} \ dx $$ Integrating by parts looks very difficult.
 A: I think this will work.
\begin{align*}
\int_{0}^{\infty} \frac{\log{x}}{x^{2}+a^{2}} \ dx &= \int_{0}^{a} \frac{\log{x}}{x^{2}+a^{2}} \ dx + \int_{a}^{\infty} \frac{\log{x}}{x^{2}+a^{2}} \ dx \\\ &= \int_{0}^{a} \frac{\log{x}}{x^{2}+a^{2}} \ dx + \int_{a}^{0} \frac{\log(a^{2}/t)}{(a^{2}/t)^{2} + a^{2}} \cdot \biggl(-\frac{a^2}{t^2}\biggr) \ dt \\\ &= \int_{0}^{a} \frac{\log{x}}{x^{2}+a^{2}} \ dx + \int_{0}^{a} \frac{2\:\log{a} - \log{t}}{t^{2}+a^{2}} \ dt \\\ &= \int\limits_{0}^{a} \frac{2\:\log{a}}{t^{2}+a^{2}} \ dt 
\end{align*}
A: First step: The change of variable $x=au$ yields
$$
I(a)=\int_0^{+\infty}\frac{\log x}{x^2+a^2}\mathrm dx=\int_0^{+\infty}\frac{\log a+\log u}{u^2+1}\frac{\mathrm du}a=\frac1a\left(\frac{\pi}2\log a+I(1)\right).
$$
Second step: Decomposing $I(1)$ into an integral on $(0,1)$ and an integral on $(1,+\infty)$ and using the change of variables $x=1/z$ in the $(0,1)$ part yields $I(1)=0$.
Conclusion: For every $a\gt0$, $I(a)=\dfrac{\pi}2\dfrac{\log a}a$.
A: The integral evaluates to $\frac{\pi\log a}{2a}.$  Here is an elementary solution.
First, we evaluate it when $a=1$.  Consider $\int_{1}^{\infty}\frac{\log u}{u^{2}+1}du$, and notice that by substituting $u=\frac{1}{v}$ we have $$\int_{1}^{\infty}\frac{\log u}{u^{2}+1}du=-\int_{0}^{1}\frac{\log v}{1+v^{2}}dv$$ and hence  $$\int_{0}^{\infty}\frac{\log u}{u^{2}+1}du=0.$$ 
Now, returning to the general case, letting $x=au$, we have $$\int_{0}^{\infty}\frac{\log x}{x^{2}+a^{2}}dx=\frac{1}{a}\int_{0}^{\infty}\frac{\log a}{u^{2}+1}du+\frac{1}{a}\int_{0}^{\infty}\frac{\log u}{u^{2}+1}du.$$  We know the second integral is $0$, and the first is $\frac{\pi}{2}$ since it is $\arctan(x)$, so we conclude that 
$$\int_{0}^{\infty}\frac{\log x}{x^{2}+a^{2}}dx=\frac{\pi\log a}{2a}.$$
Remark:  This method relied on a substitution trick.  A more general approach is to use a key-hole contour integral around the branch cut $(0,\infty)$.  
A: Let $ x=a \tan \theta,$then  $d x=a  \sec ^{2} \theta d \theta . $
\begin{aligned}
&\int_{0}^{\infty} \frac{\ln x}{x^{2}+a^{2}} d x
\\
=&\int_{0}^{\frac{\pi}{2}} \frac{\ln \left(a\tan \theta\right)}{a^{2} \sec ^{2} \theta} \cdot a \sec ^{2} \theta d \theta \\
=&\frac{1}{a} \int_{0}^{\frac{\pi}{2}} \ln (a \tan \theta) d \theta \\
=&\frac{1}{a}\left[\int_{0}^{\frac{\pi}{2}} \ln a  d \theta+\int_{0}^{\frac{\pi}{2}} \ln (\tan \theta) d \theta\right] \\
\end{aligned}
By my post,
$$\int_0^{\frac{\pi}{2}} \ln (\tan \theta) d \theta=0,$$
$$\boxed{I=\frac{\pi \ln a}{2 a }}$$
