Evaluate the improper integral $$\int_0^\infty \dfrac{\arctan(ax)-\arctan(bx)}{x}~\mathrm{d}x$$
where $a$ and $b$ are positive real numbers
I could not think of a way where to proceed from.
Please help!
 A: Hint: 
$$\arctan{a x}-\arctan{b x} = x \int_b^a \frac{dy}{1+x^2 y^2}$$
Show that you can reverse the order of integration.  You then end up integrating
$$\int_0^{\infty} \frac{dx}{1+x^2 y^2} = \frac{\pi}{2 y}$$
I assume you can handle the rest.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{%
\arctan\pars{ax}-\arctan\pars{bx} \over x}\,\dd x}:\ {\large ?}\,,
\qquad a, b\ \in {\mathbb R}\,,\quad a, b >0}$

\begin{align}
&\bbox[#ffd,5px]{\int_{0}^{\infty}{%
\arctan\pars{ax}-\arctan\pars{bx} \over x}\,\dd x}
\\[5mm] = &\
\int_{0}^{\infty}\ln\pars{x}\bracks{%
{a \over \pars{ax}^{2} + 1} -
{b \over \pars{bx}^{2} + 1}}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}{\ln\pars{x/a} \over x^{2} + 1}\,\dd x
- \int_{0}^{\infty}{\ln\pars{x/b} \over x^{2} + 1}\,\dd x
\\[5mm] = &\
\ln\pars{b \over a}\
\underbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}
_{\ds{\pi \over 2}}\
=\
\bbox[10px,border:1px groove navy]{%
\half\,\pi\,\ln\pars{b \over a}} \\ &
\end{align}
