Evaluate the improper integral

Question

$$\int_0^\infty \dfrac{\arctan(ax)-\arctan(bx)}{x}~\mathrm{d}x$$
where $a$ and $b$ are positive real numbers

I could not think of a way where to proceed from. Please help!

Answer 1

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.

Answer 2

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\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}