arctangents and logarithms are beautiful

Let:

$I(t) = \int_{0}^{\infty}\frac{\arctan(tx)}{x}dx$

Using Leibnitz rule:

$I'(t) = \int_{0}^{\infty}\frac{\partial}{\partial t} \left(\frac{\arctan(tx)}{x}\right)dx = \int_{0}^{\infty}\frac{1}{1+t^2x^2}dx$

Evaluating the very standard integral on the RHS gives:

$I'(t) = \frac{\pi}{2t} \implies I(t) = \frac{\pi}{2}\ln{t} + C$

So, to evaluate the given integral:

$A = \int_{0}^{\infty}\frac{\arctan(\pi x) - arctan(x)}{x}dx$

$A = I(\pi) - I(1)$

Substituting the obtained expression for $I(t)$ leads to:

$\boxed{A = \frac{\pi}{2}\ln{\pi} \approx 1.7981}$

Definite integrals can sometimes be simplified using iterated integrals. $\displaystyle\int_{0}^{\infty} \frac{\arctan\left(\pi x\right) -\arctan\left(x\right)}{x} dx =\displaystyle\int_{0}^{\infty} \int_{\arctan\left(x\right)}^{\arctan\left(\pi x\right) }\frac{1}{x} dy \times dx$

This the double integral is over the region enclosed within $y=\arctan\left(x\right)$ and $y= \arctan\left(\pi x\right)$

From Fubini's theorem order of iteration of definite integrals is immaterial.The region as a function of x is enclosed by:

$x=\tan\left(y\right)$ and $x=\frac{\tan\left(y\right)}{\pi}$

Integral becomes:

$\displaystyle\int_{0}^{\frac{\pi}{2}} \int_{\frac{\tan\left(y\right)}{\pi}}^{\tan\left(y\right)}\frac{1}{x} dx \times dy=\displaystyle \int_{0}^{\frac{\pi}{2}} \ln\left(x\right)\Biggr|_{\frac{\tan\left(y\right)}{\pi}}^{\tan\left(y\right)} dy$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \left(\ln\tan \left(y\right)- \ln\left(\frac{\tan\left(y\right)}{\pi} \right)\right)dy=\displaystyle \int_{0}^{\frac{\pi}{2}} \left(\ln\tan \left(y\right)-\ln\tan \left(y\right) +\ln\left(\pi\right) \right)dy$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \ln\pi dy=\frac{\pi\times\ln \pi}{2}$