Exceptional Integral

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 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 $\large I(\pi)$ .

Let,

$$ $I(b) = \displaystyle \int_{0}^{\infty} \dfrac{ \arctan bx - \arctan x}{x}dx$

$$ Differentiating under the integral symbol,

$\dfrac{\partial I}{\partial b} = \displaystyle \int_{0}^{\infty} \dfrac{x}{1 + (bx)^2} \dfrac{1}{x} dx$

$\dfrac{\partial I}{\partial b} = \displaystyle \int_{0}^{\infty} \dfrac{1}{1+(bx)^2}dx = \dfrac{\arctan bx}{b} |_{0}^{\infty}$

$\dfrac{\partial I}{\partial b} = \dfrac{\pi}{2b}$

Integrating with respect to b, $I(b) = \dfrac{\pi}{2} \times log(b)+ C$ where C is the integration constant.

$$ When $b = 1, I = 0$ , Substituting the values,

$$ $0 = \dfrac{\pi}{2} \times log(1) + C$

$$ $0 = 0 + C \rightarrow C = 0$

$I(b) = \dfrac{\pi}{2} \times log(b)$

$$ Putting $b = \pi$ we get our required integral.

$$ Required integral $= I( \pi ) = \dfrac{\pi}{2} \times log( \pi)$