TITLE: compute one improper integral involving arctangent
QUESTION [6 upvotes]: Compute $$\int_{0}^{\infty} \frac{\arctan(ax)}{x(1+x^{2})}dx$$
The answer is:
$\frac{\pi}{2}\ln(1+a)$ when $a \geq 0$
$-\frac{\pi}{2}\ln(1-a)$ when $a < 0$

Here is my problem, and I can't even dive into a appropriate first step. The most difficult part is the $\arctan$ which I have no idea to eliminate. Thank you for help.

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
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\begin{align}&\color{#c00000}{\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x}
=\half\,a\verts{a}\int_{-\infty}^{\infty}
{\arctan\pars{x} \over x\pars{x^{2} + a^{2}}}\,\dd x
\\[3mm]&=\half\,a\verts{a}\int_{-\infty}^{\infty}\
\overbrace{{\ic \over 2}\,\ln\pars{1 - \ic x \over 1 + \ic x}}
^{\ds{=\ \arctan\pars{x}}}\
{1\over x\pars{x^{2} + a^{2}}}\,\dd x
=-\,\half\,a\verts{a}\,\Im\int_{-\infty}^{\infty}
{\ln\pars{1 - \ic x} \over x\pars{x^{2} + a^{2}}}\,\dd x
\end{align}

Set $\ds{t \equiv 1 - \ic x\quad\imp\quad x = \pars{t - 1}\ic}$:
  \begin{align}&\color{#c00000}{\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x}
=-\,\half\,a\verts{a}\,\Im\int_{1 + \infty\ic}^{1 - \infty\ic}
{\ln\pars{t} \over \pars{t - 1}\ic\bracks{-\pars{t - 1}^{2} + a^{2}}}\,\ic\,\dd t
\\[3mm]&=-\,\half\,a\verts{a}\,\Im\color{#00f}{%
\int_{1 - \infty\ic}^{1 + \infty\ic}{\ln\pars{t}\over
\pars{t - 1}\pars{t - r_{-}}\pars{t - r_{+}}}\,\dd t}
\quad \mbox{where}\quad r_{\pm} = 1 \pm \verts{a}
\end{align}

Now, we'll evaluate the $\ds{\color{#00f}{\mbox{'blue integral'}}}$.
We take the "$\ds{\ln}$-branch cut" along the negative $\ds{t}$-semi-axis and close the contour in a semi-circle "to the right" $\ds{\pars{~t > 1~}}$: 
\begin{align}&\color{#c00000}{\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x}
=-\,\half\,a\verts{a}\,\Im\bracks{-2\pi\ic\,
{\ln\pars{r_{+}} + \ic 0 \over \pars{r_{+} - 1}\pars{r_{+} - r_{-}}}}
\\[3mm]&=\pi\,a\verts{a}\,\bracks{%
{\ln\pars{1 + \verts{a}} \over \verts{a}\pars{2\verts{a}}}}
\end{align}

$$
\color{#66f}{\large\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x
={\pi \over 2}\,\sgn\pars{a}\ln\pars{1 + \verts{a}}}
$$