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: First, decompose the denominator in partial fractions and perform a change of variable which leads to the integral of  Cos[x] / x.
Please, have a look at http://en.wikipedia.org/wiki/Trigonometric_integral#Sine_integral
Then the antiderivative of the integrand is just
1/2 (SinIntegral[Pi (1 + t) / 2] - SinIntegral[Pi (1 - t) / 2])
and the integration  between 0 and infinity leads to Pi / 2.   
A: Define
$$f(z)=\frac{e^{\pi iz/2}}{1-z^2}\;,\;\;C_R:=[-R,-1-\epsilon]\cup\gamma_{-1,\epsilon}\cup[1+\epsilon,\epsilon]\cup\gamma_{1\epsilon}\cup[1+\epsilon,R]\cup\Gamma_R$$
with $\;\epsilon, R\in\Bbb R^+\;$ and
$$\gamma_{r,s}:=\{r+se^{it}\;;\;0\le t\le \pi\}\;,\;r,s\in\Bbb R^+\;,\;\Gamma_R:=\{Re^{it}\;;\;0\le t\le \pi\}\;$$
Now, as the two poles of the function are simple:
$$\begin{align*}\text{Res}_{z=-1}(f)&=\lim_{z\to -1}(z+1)f(z)=\frac{e^{-\pi i/2}}{2}=-\frac i2\\
\text{Res}_{z=1}(f)&=\lim_{z\to 1}(z-1)f(z)=-\frac{e^{\pi i/2}}{2}=-\frac i2\end{align*}$$
so using the corollary to the lemma in the second answer here (watch the direction of the integration!), and the residue theorem for integration, and using also Jordan's Lemma, we get:
$$0=\lim_{R\to\infty\,,\,\epsilon \to 0}\oint\limits_{C_R}f(z) dz=\int\limits_{-\infty}^\infty f(x)dx+\pi i(i)\implies$$
$$\int\limits_{-\infty}^\infty \frac{e^{\pi ix/2}}{1-x^2}dx=\pi$$
Now just compare real parts in both sides and take into account your integrand function is an even one...
A: 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

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

