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.
 A: If you make a first integration by parts $u=\sin^4(x)$, $v'=\frac {dx}{x^2}$, $u'=4\sin^3(x)\cos(x)dx$, $v=-\frac 1x$, you obtain $$\int \frac{\sin^4 (x)}{x^2}dx=-\frac{\sin^4(x)}x+\int\frac{4\sin^3 (x)\cos(x)}{x}dx$$ Now $$A=4\sin^3 (x)\cos(x)=\Big(2\sin(x)\cos(x)\Big)\Big(2\sin^2(x)\Big)=\sin(2x)(1-\cos(2x))$$ $$A=\sin(2x)-\frac 12 \sin(4x)$$ So, $$I=\int\frac{4\sin^3 (x)\cos(x)}{x}dx=\int\frac{\sin(2x)}x dx-\frac 12\int\frac{\sin(4x)}x dx$$ Changing variables $2x=y$ in the first integral and $4x=y$ in the second integral simplifies to $$I=\frac 12\int\frac{\sin(y)}y dy=\frac 12\text{Si}(y)$$ Now, use the bounds : the first term is $0$ and what is left is then half the value of the sine integral $\text{Si}(y)$ for an infinite value of $y$ and this last is $\frac \pi 2$ (refer to the post mentioned by Travis).
Edit
Concerning $\int\frac{\sin(ax)}x dx$, make a change of variable $ax=y$ to get $$\int\frac{\sin(ax)}x dx=\int\frac{\sin(y)}y dy=\text{Si}(y)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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}
A: $$
\begin{align}
\int_0^\infty\frac{\sin^4(x)}{x^2}\,\mathrm{d}x
&=4\int_0^\infty\frac{\sin^3(x)\cos(x)}{x}\,\mathrm{d}x\tag{1}\\
&=\frac12\int_0^\infty\frac{2\sin(2x)-\sin(4x)}{x}\,\mathrm{d}x\tag{2}\\[3pt]
&=\frac\pi2-\frac\pi4\tag{3}\\[6pt]
&=\frac\pi4\tag{4}
\end{align}
$$
Explanation:
$(1)$: integrate by parts
$(2)$: write $\sin^3(x)=\frac{3\sin(x)-\sin(3x)}4$, then $\sin(ax)\cos(x)=\frac{\sin((a+1)x)+\sin((a-1)x)}2$
$(3)$: use $\int_0^\infty\frac{\sin(ax)}{x}\,\mathrm{d}x=\frac\pi2$
A: I think this question was asked the other week, I will look for a link (and thus remove this answer if I find it). 
In the mean time, let me provide a way to calculate this integral in term of $\int_0^{+\infty}(\sin x)/x\,dx$. I don't know if it fits as an answer, but it is too long for being a comment.
Step 1 Use $\cos^2x+\sin^2x=1$ and the formula $\sin 2x=2\sin x\cos x$ to reduce the problem to calculate the integral
$$
\int_0^{+\infty}\frac{\sin^2x}{x^2}.
$$
(You don't get exactly this integral, but a sum of these kinds of integrals.)
Step 2 Integrate by parts, where you move the derivative from $-1/x$ to $\sin^2x$. This will give you something like $\int_0^{+\infty}(\sin x)/x\,dx$. I leave the details to you.
A: 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}
