real analysis - 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?

Answer

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

$$
L(m)=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}.
$$