Integrating Difference Of Arctangents?

Relevant wiki: Differentiation Under the Integral Sign

Let,

$I(a)= \int_{0}^{\infty} \frac{\tan^{-1}(ax) - \tan^{-1}(x)}{x}dx$

$I'(a)= \int_{0}^{\infty} \frac{dx}{1+a^2x^2} = \frac{\pi}{2a}$

$\text{Now, I(1) = 0}, thus$

$I(\pi) = I(1) + \int_{1}^{\pi} I'(a)da$

$= \frac{\pi \ln {\pi}}{2}$

$a= \pi, b = \pi, c = 2$

$a-b+c = 2$

This problem is in integration tricks

Here is @Anastasiya Romanova 's solution

One may utilize Frullani's theorem to evaluate the integral. $\int_0^\infty \frac{f(ax)-f(bx)}{x}\,dx=\lim_{t\to\infty}\bigg(f(0)-f(t)\bigg)\ln\left(\frac{b}{a}\right)$ Hence our integral is simply $\frac{\pi}{2}\ln\pi$ One may also use Feynman's trick (which is developed in @Vatsalya Tandon ' solution) by considering $I(a)=\int_0^\infty \frac{\arctan ax-\arctan x}{x}\,dx$ so that $\large I(1)=0$ . After differentiating w.r.t. $\large a$ and then integrating it back, one may obtain $I(a)=\frac{\pi}{2}\ln a$ and our considered integral is $\large I(\pi)$ .

$\large \int_{0}^{\infty} \dfrac{\tan^{-1}(\pi x) - \tan^{-1}(x)}{x}\, dx$ We can write the given integral as, $\large \int_{0}^{\infty}\large \int_{x}^{\pi x} \dfrac{1}{(1+y^2)x}\,dydx$ Changing the order of integration,we get $\large \int_{0}^{\infty}\large \int_{y/\pi}^{y} \dfrac{1}{(1+y^2)x}\,dxdy$ $=\large \int_{0}^{\infty}\dfrac{\ln(\pi)}{1+y^2}\,dy$ $=\frac{\pi ln(\pi)}{2}$