(Edit: Thank you Vladimir for providing the references for the closed form value of the integrals. My revised question is then to how to derive this closed form.)
For all $n\in\mathbb{N}^+$, define $\mathcal{I}_n$ by the definite integral,
$$\mathcal{I}_n:=\int_{0}^{\infty}\frac{\sin^n{(x)}}{x^n}\mathrm{d}x.$$
Prove that $\mathcal{I}_n$ has the following closed form:
$$\mathcal{I}_n\stackrel{?}=\pi\,2^{-n}\left(n\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{(-2)^k(n-2k)^{n-1}}{k!(n-k)!}\right),~~\forall n\in\mathbb{N}^+.$$
Integrals of small positive integer powers of the $\operatorname{sinc}$ function come up on a regular basis here, but it occurred to me that while I probably know the derivations for the $1\le n\le 4$ cases like the back of my hand, I can't recall ever working the integrals wfor any value of $n$ higher than that. The values of the first four integrals are,
$$\mathcal{I}_1=\frac{\pi}{2},\\
\mathcal{I}_2=\frac{\pi}{2},\\
\mathcal{I}_3=\frac{3\pi}{8},\\
\mathcal{I}_4=\frac{\pi}{3}.$$
So I set out to first calculate $\mathcal{I}_5$ to see if any obvious pattern jumped out (and see if the trend of being equal to rational multiples of $\pi$ continued). I wound up getting frustrated and asking WolframAlpha instead. It turns that while the first four cases hinted very much at the possibility of a simple pattern relating the values of $\mathcal{I}_n$ for different positive integers $n$ (or possibly two separate patterns for even and odd $n$), the next few values most definitely did not:
$$\mathcal{I}_5=\frac{115\pi}{384},\\
\mathcal{I}_6=\frac{11\pi}{40},\\
\mathcal{I}_7=\frac{5887\pi}{23040}\\
\mathcal{I}_8=\frac{151\pi}{630}.$$
So my questions are, 1) is there a systematic way to compute these integrals for all $n$?; and 2) is there an elegant way to represent these values in closed form for general $n$?
Best Answer
All sinc function integrals of the type
$$I_n=\int_{0}^{\infty}\frac{\sin^n{(x)}}{x^n}\mathrm{d}x$$
can be expressed using the following general form:
$$\displaystyle \int_{0}^{\infty}\frac{\sin^a{(x)}}{x^b}\mathrm{d}x=\frac{(-1)^{\lfloor(a-b)/2 \rfloor} \cdot \pi^ {1-c}}{ 2^{a-c}(b-1)!}\sum_{k=0}^{\lfloor a/2-c \rfloor } (-1)^k { a \choose k}(a-2k)^{b-1} Log^c(a-2k) $$
where $a$ and $b$ and are positive integers, $c\equiv (a-b) \pmod 2$, and $\lfloor j \rfloor $ denotes the floor function. When $a=b=n$, then $c=0$ and the equation simplifies in
$$\displaystyle I_n=\frac{ \pi}{ 2^{n}(n-1)!}\sum_{k=0}^{\lfloor n/2 \rfloor }(-1)^k { n \choose k}(n-2k)^{n-1} $$
so that you can obtain the rational coefficients for each $n$ by dividing the last expression to $\pi$. This gives you the sequence 1/2, 1/2, 3/8, 1/3, 115/384, 11/40, 5887/23040, 151/630, 259723/1146880, 15619/72576, 381773117/1857945600, 655177/3326400.....
The last equation can also be further simplified in
$$\displaystyle I_n=\frac{n \pi}{ 2^{n}}\sum_{k=0}^{\lfloor n/2 \rfloor } \frac{(-1)^k (n-2k)^{n-1}}{k!(n-k)!} $$