How to integrate $\int_0^{\infty}\left(\frac{\sin(x)}{x}\right)^m dx$?

Question

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 1

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

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

Answer 2

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 $\pi$.

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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

$$\int_0^{\infty}\left(\frac{\sin x}x\right)^mdx=\frac{\pi}{2^{m}(m-1)!}\sum_{k=0}^{\left \lfloor{\frac{m}{2}}\right \rfloor }(-1)^{k}\binom{m}{k}(m-2k)^{m-1}$$