How to compute the integral of $\log x/(x+17)^2$ , x from 0 to $\infty$, using contour integration? Thanks! How to compute the integral of $\log x/(x+17)^2$ , x from 0 to $\infty$, using contour integration? 
$$\int_{0}^\infty \frac{\log x}{(x+17)^2}dx$$
Thanks!
 A: Contour Integration
Let $\gamma=[0,Re^{i\epsilon}]\cup Re^{i[\epsilon,2\pi-\epsilon]}\cup[Re^{-i\epsilon},0]$. The integral over the arc vanishes and the contour circles one pole at $x=-17$.
$\hspace{3.5cm}$
$$
\begin{align}
\int_\gamma\frac{\log(z)^2}{(z+17)^2}\,\mathrm{d}z&=\int_0^\infty\frac{\log(x)^2-(\log(x)+2\pi i)^2}{(x+17)^2}\,\mathrm{d}x\\
&=\int_0^\infty\frac{-4\pi i\log(x)+4\pi^2}{(x+17)^2}\,\mathrm{d}x\\
\end{align}
$$
The residue of $\frac{\log(z)^2}{(z+17)^2}$ at $z=-17$ is $\frac{2\log(17)+2\pi i}{-17}$, so the integral above must be
$$
\frac{4\pi^2-4\pi i\log(17)}{17}
$$
Taking the imaginary parts we get
$$
\int_0^\infty\frac{\log(x)}{(x+17)^2}\,\mathrm{d}x=\frac{\log(17)}{17}
$$

Integration by Parts
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{(x+17)^2}\,\mathrm{d}x
&=\lim_{a\to0}\int_a^\infty\log(x)\,\mathrm{d}\frac{-1}{x+17}\\
&=\lim_{a\to0}\left(\int_a^\infty\frac1{x(x+17)}\,\mathrm{d}x
-\left[\frac{\log(x)}{x+17}\right]_a^\infty\right)\\
&=\lim_{a\to0}\left(\frac1{17}\log(a+17)\color{#C00000}{-\frac1{17}\log(a)+\frac{\log(a)}{a+17}}\right)\\
&=\frac{\log(17)}{17}\color{#C00000}{-\lim_{a\to0}\frac{a\log(a)}{17(a+17)}}\\
&=\frac{\log(17)}{17}
\end{align}
$$
