How to integrate $\int_0^{\infty}\left(\frac{\sin(x)}{x}\right)^m dx$? How to evaluate $L(m):=\int_0^{\infty}\left(\frac{\sin(x)}{x}\right)^m dx$? 
I am familiar with the case $m=1$, but what about the general one?
 A: I recently explained this issue to a student in the following way:
Let $f(x)=\sin^m(x)$ where $m>1$. Integration by parts $m-1$ times gives the formula 
$$
\int_0^{\infty}\frac{f(x)}{x^m}dx=\frac{1}{(m-1)!}\int_0^{\infty}\frac{f^{(m-1)}(x)}{x}dx.
$$
Now we have the formulae 
\begin{align}
\sin^{2n}(x)&=\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^{n-k}\binom{2n}{k}\cos((2n-2k)x)+\frac{1}{2^{2n}}\binom{2n}{n} \\
\sin^{2n-1}(x)&=\frac{1}{2^{2n-2}}\sum_{k=0}^{n-1}(-1)^{n-k-1}\binom{2n-1}{k}\sin((2n-2k-1)x).
\end{align}
Differentiate the former $2n-1$ times and the latter $2n-2$ times to obtain with the very first equation 
\begin{align}
L(2n)&=\frac{\pi}{2^{2n}(2n-1)!}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n}{k}(2n-2k)^{2n-1} \\
L(2n-1)&=\frac{\pi}{2^{2n-1}(2n-2)!}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{k}(2n-2k-1)^{2n-2},
\end{align}
since
$$
\int_0^{\infty}\frac{\sin((2n-2k)x)}{x}dx=\int_0^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}.
$$
This yields in a campact shape 
$$
L(m)=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}.
$$
A: In general you can substitute the $x^{-m}$-term by using the Laplace transform. To be precise by using 
$$\mathcal{L}\{t^n\}(x)=\int_0^{\infty}t^ne^{-xt}~dt\stackrel{n\in\mathbb{N}}{=}\frac{n!}{x^{n+1}}\\
\Leftrightarrow\frac1{x^{n+1}}=\frac1{n!}\int_0^{\infty}t^ne^{-xt}~dt$$
the integral becomes for $n+1=m$
$$\begin{align}
\int_0^{\infty}\left(\frac{\sin x}{x}\right)^mdx&=\int_0^{\infty}\sin^m x\left[\frac1{(m-1)!}\int_0^{\infty}t^{m-1}e^{-xt}~dt\right]dx\\
&=\frac1{(m-1)!}\int_0^{\infty}t^{m-1}\int_0^{\infty}e^{-xt}\sin^m x~dx~dt
\end{align}$$
where the inner integral can be evaluated by applying the trigonometric identities repeatedly
$$\sin^{2n}x=\frac{1-\cos(2nx)}{2}\text{ and }\cos^{2n}x=\frac{1+\cos(2nx)}{2}$$
and for the remaining combined sine-cosine terms one might recall the double-angle formula. Some reshaping leads to 
$$\sin(n)\cos(m)=\frac12(\sin(n+m)+\sin(n-m))$$
By using these the whole inner integral can be broken down to the simple standard integrals of the form
$$\begin{align}
\int e^{ax}\sin x~dx\text{ and }\int e^{ax}\cos x~dx
\end{align}$$
and the outer integral will remain in the form
$$\int\frac{1}{a^2+t^2}~dt=\frac1a\arctan\left(\frac ta\right)$$
from hereon it can be evaluated directly in terms of $\pi$.

By doing the whole process in a more general way $($i.e. the sine power differs from the algebraic power$)$ one may arrive at formula $(37)$ here which also provides the special case for matching powers given by
$$\int_0^{\infty}\left(\frac{\sin x}x\right)^mdx=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}$$
