I want to evaluate
$$\int_{0}^{\infty}\frac{\sin(\pi x)}{1 + x^2}$$
by the residue theorem.
This function is not even, so I can not use the semi-circle approach.
I was trying to integrate over the triangle $T= \{(0,0),(T,0),(T,T)\}$ and then take the limit as $T\to\infty$ but I had no luck with the integrals.
I parametrize each edge of the triangle and I know that the function is analytic on T and the contour integral is 0. I dont know how to deal with the remaining integrals over the sides of T.
Without contour integration.
As Simply Beautiful Art commented, writing
$$\frac{\sin(\pi x)}{1+x^2}=\frac{\sin(\pi x)}{(x+i)(x-i)}=\frac i2 \left(\frac{\sin(\pi x)}{x+i}- \frac{\sin(\pi x)}{x-i}\right)$$ and using $$\int \frac{\sin (a x)}{x+b} \,dx=\cos (a b)\, \text{Si}(a(x+ b))-\sin (a b)\, \text{Ci}(a(x+ b))$$ $$\int_0^\infty \frac{\sin (a x)}{x+b} \,dx=\frac{a (2 \text{Ci}(b |a|) \sin (b |a|)+\cos (a b) (\pi -2 \text{Si}(b |a|)))}{2 |a|}$$ we should end with $$\int_0^\infty \frac{\sin(\pi x)}{1+x^2}\,dx=\text{Shi}(\pi ) \cosh (\pi )-\text{Chi}(\pi ) \sinh (\pi )$$ where appear the hyperbolic sine and cosine integral functions.