Compute $\int_0^{\infty}\frac{\cos(\pi t/2)}{1-t^2}dt$ Compute $$\int_0^{\infty}\frac{\cos(\pi t/2)}{1-t^2}dt$$

The answer is $\pi/2$. The discontinuities at $\pm1$ are removable since the limit exists at those points.

A related problem. Follow the steps
i) Use partial fraction 

$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{\cos(\pi t/2)}{1-t^2}dt =\frac{1}{4}\int_{-\infty}^{\infty}\frac{\cos(\pi t/2)}{t+1}dt + \frac{1}{4}\int_{\infty}^{\infty}\frac{\cos(\pi t/2)}{1-t}dt. $$

ii) Use the change of variables $1+t=u$ and $1-t=u$ for the integrals on the RHS.
iii) Use the result

$$ \int_{-\infty}^{\infty} \frac{\sin(y)}{y}dy=\pi. $$

