# Closed form for integral of integer powers of Sinc function

(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 $n$?

• AFAIK, no simple formula exists for these rational coefficients. See oeis.org/A049330, oeis.org/A049331, R. Baillie, D. Borwein and J. M. Borwein, Surprising Sinc Sums and Integrals, Amer. Math. Monthly, 115 (2008), 888-901. Commented Aug 12, 2014 at 3:00
• Actually, there are some formulae. See (37), (38) at MathWorld. Commented Aug 12, 2014 at 3:09
• @VladimirReshetnikov Ah, thank you. I don't know why it didn't occur to me to check there first. Also, just after you posted that I stumbled across almost the same formula at the bottom of this page on the Wolfram Functions site. Commented Aug 12, 2014 at 3:16
• There is the full text of that paper at J. M. Borwein's page at carma.newcastle.edu.au/jon/sinc-sums.pdf Commented Aug 12, 2014 at 3:33
• @DavidH Did you see this math.stackexchange.com/questions/307510/… ?
– user146010
Commented Aug 12, 2014 at 14:22

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

$$\displaystyle I_n=\frac{n \pi}{ 2^{n}}\sum_{k=0}^{\lfloor n/2 \rfloor } \frac{(-1)^k (n-2k)^{n-1}}{k!(n-k)!}$$

• Yes, these are the formulas that Vladimir Reshetnikov cited in the comments above. If you can provide a derivation of these formulas, I will accept your answer. Commented Aug 12, 2014 at 4:15
• Hi David, you can find an elegant derivation of this formula, based on Fourier transforms, in an answer of Nick Strehlke to a previous post at math.stackexchange.com/questions/331404/…. Commented Aug 12, 2014 at 4:37
• The approach by Strehlke seems to me a nice and relatively simple method of derivation. Commented Aug 12, 2014 at 4:41

Here's an idea for even n; I haven't looked at any of the links above, so hope this is not the same as that in Strehlke. [Ok, after finishing this, took a look at Strehlke. Approach below is not the same as his.]

The sinc function is the Fourier transform of a boxcar. Here I'll work with a boxcar that is 1 for x in -1/2,1/2 and zero otherwise. The width of the boxcar will relate to the frequency of the sinc, but one can always rescale the integral given so that the frequency leads to an inverse FT with the boxcar I use here.

So the integral can be written as $\int (\frac{\sin (\omega x)}{\omega x})^ndx=\int \Pi_{i=1}^nFT\{boxcar*boxcar\cdots * boxcar \}$, where * is convolution. This is using the Convolution Theorem, so that the FT of a convolution is the product of the FTs.

Now, we can evaluate the convolutions pairwise (this is why I needed n even). boxcar * boxcar = triangle, where for the particular boxcars I chose, the slope of the triangle's legs is $\pm 1$. I.e., the triangle is at the origin, looking like /\ with lines of slope $\pm 1$. So we now have convolutions of n/2 triangles like this.

The overlap area of these 2 triangles is $A_{overlap}=\frac{1}{2}(2-\Delta)$ for $\Delta\leq 2$ and zero otherwise, where $\Delta$ is the offset in the convolution. When the triangles sit on top of each other, the overlap is just 1, and when they are not overlapping at all, $\Delta>2$, the overlap is zero.

So we can see that the convolution of 2 triangles is a new triangle whose legs, /\, have slope $\pm 1/2$.

Triangles are in this sense eigenvectors of the convolution operator. So we had n/2 triangles with slope $\pm 1$ from convolving the boxcars, and now we have n/4 triangles with slope $\pm 1/2$ from this second round of convolutions. Convolving again brings us to n/8 triangles with slopes $\pm 1/4$. In general, convolving a triangle with itself divides the slope by 2. So here, the slope is $1/2^c$, c the number of convolutions of triangles (not counting that we did on the boxcars), and the number of triangles $n/2^{c+1}$. To get down to one triangle, $c=\ln n/\ln 2 -1$. It will have slope $m=\frac{1}{2^{\ln n/\ln 2-1}}$. We can then take its Fourier transform as

$\int_{-1/m}^0 e^{ikx} (mk+1) dk-\int_{0}^{1/m} e^{ikx} (1-mk) dk$, where I've split the integral to reflect the piecewise definition of a triangle; note that the bounds also reflect the triangle shape.

Note the triangle is symmetric about the origin, so only a cosine will contribute to the FT; the sine's contribution cancels off. This integral can be done easily and gives a simple function of x that can be integrated from 0 to infinity.

Also, note that this approach will still work for odd n, as we would just split the boxcars convolutions into an even number of boxcars convolved with each other, then convolved with one more boxcar. So if the power is now n+1, we could just take the final triangle from power n case reduction, and then convolve it with a boxcar---also a simple function. Then FT that and integrate over x.

• First of all, your formula $\int (\frac{\sin (\omega x)}{\omega x})^ndx$=$\int \Pi_{i=1}^nFT\{boxcar*boxcar\cdots*boxcar \}$ should be $\int (\frac{\sin (\omega x)}{\omega x})^ndx=\int \Pi_{i=1}^n FT\{boxcar\}$. Then, it's very difficult to follow your argumentation without formulas. My experience with Fourier Transform is that, as long as you have not written down all the calculation steps you are never sure that there is no (more or less) subtile flaw..., one of them being the extension of $\int fg = \int \hat f \hat g$ to more than two functions. Commented Apr 24, 2016 at 21:55