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The following question comes from Some integral with sine post $$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$ but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!


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This is related:… and this is too:… – 1015 Feb 18 '13 at 22:27
I think someone made a paper about this, but can't recall a source. – Pedro Tamaroff Feb 18 '13 at 22:37
@PeterTamaroff You must be referring to the "Surprising sinc sums and integrals" by Baillie, Borwein and Borwein. ( pdf ) – Sasha Feb 18 '13 at 22:50
Note that the integral has an interpretation of $2 f_{X_n}(0)$, where $X_n$ is the sum of $n$ uniform on $(-1,1)$ random variables, and $f_X(x)$ denotes the pdf. Using the central limit theorem, one can find large $n$ asymptotics, $\int_0^\infty \sin^n(x)/x^n \mathrm{d} x \approx \sqrt{\frac{3 \pi}{2 n}} $ – Sasha Feb 18 '13 at 23:00
@Sasha Yes, indeed! Thank you. – Pedro Tamaroff Feb 18 '13 at 23:01
up vote 17 down vote accepted

Here's another approach.

We have $$\begin{eqnarray*} \int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \int_{-\infty}^\infty dx\, \left(\frac{\sin x}{x-i\epsilon}\right)^n \\ &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \int_{-\infty}^\infty dx\, \frac{1}{(x-i\epsilon)^n} \left(\frac{e^{i x}-e^{-i x}}{2i}\right)^n \\ &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \frac{1}{(2i)^n} \int_{-\infty}^\infty dx\, \frac{1}{(x-i\epsilon)^n} \sum_{k=0}^n (-1)^k {n \choose k} e^{i x(n-2k)} \\ &=& \lim_{\epsilon\to 0^+} \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^n (-1)^k {n \choose k} \int_{-\infty}^\infty dx\, \frac{e^{i x(n-2k)}}{(x-i\epsilon)^n}. \end{eqnarray*}$$ If $n-2k \ge 0$ we close the contour in the upper half-plane and pick up the residue at $x=i\epsilon$. Otherwise we close the contour in the lower half-plane and pick up no residues. The upper limit of the sum is thus $\lfloor n/2\rfloor$. Therefore, using the Cauchy differentiation formula, we find $$\begin{eqnarray*} \int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n &=& \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} \frac{2\pi i}{(n-1)!} \left.\frac{d^{n-1}}{d x^{n-1}} e^{i x(n-2k)}\right|_{x=0} \\ &=& \frac{1}{2} \frac{1}{(2i)^n} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} \frac{2\pi i}{(n-1)!} (i(n-2k))^{n-1} \\ &=& \frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}. \end{eqnarray*}$$ The sum can be written in terms of the hypergeometric function but the result is not particularly enlightening.

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+1 for reminding me of the Irwin-Hall distribution. The sum you derives is exactly the $2 f_X(0)$, where $X$ is the Irwin-Hall random variable. – Sasha Feb 19 '13 at 1:09
@Sasha: Indeed it appears that the integral is $\frac{\pi}{2}f_X(\frac{n}{2},n)$. Thank you for informing me of this interesting connection! – user26872 Feb 19 '13 at 2:11
@oen: good shot! Long time I haven't seen you around! So, welcome! :-) (+1) – user 1618033 Feb 19 '13 at 9:33
@Chris'ssisterandpals: Glad to help. Your questions often have connections to more areas of mathematics than it appears at first glance, which make them very interesting indeed. – user26872 Feb 19 '13 at 11:58

Just to verify oen's post (since there is a post with a different answer), I will post the answer I got.

$|\sin(z)|\le e^{|\mathrm{Im}(z)|}$; therefore, on the strip $|\mathrm{Im}(z)|\le1$, we have $|\sin(z)|\le e$. Thus, $\left(\frac{\sin(z)}{z}\right)^n$ vanishes as $|z|\to\infty$ in that strip and therefore, $$ \int_{-\infty}^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z =\int_{-\infty-i}^{\infty-i}\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z\tag{1} $$ Next define two contours $\gamma^+$ and $\gamma^-$. $\gamma^+$ goes from $-R-i$ to $R-i$ then circles back through the upper half plane along $|z+i|=R$. $\gamma^-$ goes from $-R-i$ to $R-i$ then circles back through the lower half plane along $|z+i|=R$.

Using the binomial theorem, we get $$ \left(\frac{\sin(z)}{z}\right)^n=\frac1{(2iz)^n}\sum_{k=0}^n(-1)^k\binom{n}{k}e^{(n-2k)iz}\tag{2} $$ Integrate the terms where $n-2k\ge0$ along $\gamma^+$ and the others along $\gamma^-$. Since $\gamma^-$ doesn't enclose any singularities, we can ignore that integral. Therefore, $$ \begin{align} \int_0^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z &=\frac12\int_{\gamma^+}\frac1{(2iz)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}e^{(n-2k)iz}\mathrm{d}z\\ &=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\mathrm{Res}\left(\frac{e^{(n-2k)iz}}{z^n},0\right)\\ &=\frac{\pi i}{(2i)^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\frac{(n-2k)^{n-1}i^{n-1}}{(n-1)!}\\ &=\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1}\tag{3} \end{align} $$

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I'll write $I = \int_{-\infty}^{\infty} \left(\frac{\sin z}{z} \right)^n dz$

First, to simplify matters let's take $n$ odd and $\geq 3$. Let $C_{\epsilon}^+$ be the contour along the real line that takes a semicircular detour into the upper half plane about the origin, and let $C_{\epsilon}^-$ be the same for the lower half plane. We use continuity of the integrand to argue that $$ I = \lim_{\epsilon \rightarrow 0} \int_{C_{\epsilon}^{\pm}} = \frac{1}{2} \lim_{\epsilon \rightarrow 0} \left( \int_{C_{\epsilon}^+} + \int_{C_{\epsilon}^-} \right) $$ Now think about $(\sin x)^n$: it's a sum of exponential terms of the form $e^{i l x}$ for $-n \leq l \leq n$ with some coefficients. You should convince yourself that any $l < 0$ term is killed by $\int_{C_{\epsilon}^-}$ and any $l > 0$ term is killed by $\int_{C_{\epsilon}^+}$. Moreover by completing these contours with large semicircles, you can derive ($l > 0$): $$ \int_{C_{\epsilon}^{\mp}} \frac{e^{\pm i l x}}{x^n} dx = \mp 2 \pi i \frac{(\pm i l)^{n-1}}{(n-1)!} $$ Summing everything up and noticing that there is no $\epsilon$ dependence, and keeping track of signs (which I failed to do on a first pass) we've shown that, $$ I = \frac{\pi }{2^{n-1} (n-1)!} \sum_{l = 0}^{(n-1)/2} (-1)^{n-1-l}\left(\begin{array}{c}n \\ l \end{array} \right) (n-2l)^{n-1} $$ I hope that wasn't too much (or too little).

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You've written $I$ to be twice the asked integral. However, when $n=2$, your answer gives $-\pi$ as the integral of a positive function. The problem seems to be that the exponent of $-1$ has an extraneous $n-1$. – robjohn Feb 19 '13 at 11:05
@robjohn: Since $n$ is odd by assumption $(-1)^{n-1} = 1$. (+1) to your different approach. – user26872 Feb 19 '13 at 12:05
@oen: Ah, I hadn't noticed that this answer was restricted to odd $n$. That assumption doesn't seem to be used in any part of the proof. Thanks for the upvote; I made my answer CW in hopes that it would not steal any votes. – robjohn Feb 19 '13 at 12:14
@robjohn: That is very considerate, but your answer deserves upvotes! – user26872 Feb 19 '13 at 12:27
@oen: I just looked more closely at your answer and I see it is a different approach from mine. You changed the integrand where I changed the contour. Perhaps I will remove the CW :-) – robjohn Feb 19 '13 at 12:37

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