How do I find $\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$? I need help evaluating the following integral
$$\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$$
which should probably be equal to $\frac{\pi}{4}$
Using some trigonometric manipulations I got $\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8}$ which using integration by parts doesn't lead me to anything pretty.
Update: Not sure if I should post this as a separate question but getting explanation why $\int_{0}^{\infty} \frac{\sin{ax}}{x} = \frac{\pi}{2}$ for a positive integer $a$ could help me solve this question.

Another way to solve this problem is to use parametrized integral: Let
$$ I(s)=\int_0^\infty e^{-sx}\frac{\sin^4x}{x^2}dx. $$
Then
\begin{eqnarray}
I''(s)&=&\int_0^\infty e^{-sx}\sin^4xdx\\
&=&\int_0^\infty e^{-sx}(\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8})dx\\
&=&\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(s^2+16)}
\end{eqnarray}
so 
\begin{eqnarray}
I(0)&=&\int_0^\infty\int_{t}^\infty(\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(s^2+16)})dsdt\\
&=&\frac{1}{16}\int_0^\infty(-6\ln t+4\ln(t^2+4)-\ln(t^2+16))dt\\
&=&\left.-\frac{3}{8} t \ln t+\frac{1}{4} t \ln \left(t^2+4\right)-\frac{1}{16} t \ln
   \left(t^2+16\right)-\frac{1}{2} \arctan
   ^{-1}\left(\frac{t}{4}\right)+\arctan ^{-1}\left(\frac{t}{2}\right)\right|_0^\infty\\
&=&\frac{\pi}{4}.
\end{eqnarray}
