TITLE: How do I find $\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$?
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{2}}\,\dd x & =
{1 \over 2}\int_{-\infty}^{\infty}{\sin^{4}\pars{x} \over x^{2}}\,\dd x =
{1 \over 2}\lim_{N \to \infty}\int_{-N\pi}^{N\pi}{\sin^{4}\pars{x} \over x^{2}}\,\dd x
\\[5mm] & =
{1 \over 2}\lim_{N \to \infty}\sum_{k = -N}^{N - 1}\int_{k\pi}^{k\pi + \pi}{\sin^{4}\pars{x} \over x^{2}}\,\dd x =
{1 \over 2}\lim_{N \to \infty}\sum_{k = -N}^{N - 1}\int_{0}^{\pi}{\sin^{4}\pars{x} \over \pars{x + k\pi}^{2}}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\pi}\sin^{4}\pars{x}\csc^{2}\pars{x}\,\dd x =
{1 \over 2}\int_{0}^{\pi}{1 - \cos\pars{2x} \over 2}\,\dd x = \bbx{\pi \over 4}
\end{align}