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!!!


user 1591719
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  • This is related: http://math.stackexchange.com/questions/260574/how-the-calculate-int-0-infty-frac-sin2-xx2-mathrmd-x?rq=1 and this is too: http://math.stackexchange.com/questions/13344/proof-for-an-integral-involving-sinc-function – Julien Feb 18 '13 at 22:27
  • I think someone made a paper about this, but can't recall a source. – Pedro Feb 18 '13 at 22:37
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    @PeterTamaroff You must be referring to the "Surprising sinc sums and integrals" by Baillie, Borwein and Borwein. ( [pdf](https://web.cs.dal.ca/~jborwein/sinc-sums.pdf) ) – Sasha Feb 18 '13 at 22:50
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    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 Feb 18 '13 at 23:01
  • @Sasha: What an astounding paper! :-) – joriki Feb 18 '13 at 23:25
  • @Sasha: Could you please elaborate on the detailed derivation relating the central limit theorem to this integral? – Hans Dec 12 '18 at 12:08

5 Answers5


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](http://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_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 01:09
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    @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 02:11
  • @oen: good shot! Long time I haven't seen you around! So, welcome! :-) (+1) – user 1591719 Feb 19 '13 at 09:33
  • @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 know this is a 5 year old answer but I am wondering why you chose this contour over a circular or semicircular one, for example. I know how to do the case $n=1$ with semicircular contours, but it wouldn't immediately come to me why not to use the same contour for arbitrary $n.$ Really apreciate it if you can clarify your thought process here, this is such a clean answer. – Hobbyist May 13 '18 at 11:12
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    @Hobbyist: This contour *is* a semi-circular contour, simply shifted by $-i$. Since the integrand has no (non-removable) singularities and $\frac{\sin(x)}x$ vanishes at $\pm\infty$, shifting $(-\infty,\infty)$ to $(-\infty-i,\infty-i)$ does not change the integral. This shift then makes the manipulation above much "cleaner" because, when we separate the exponentials in $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, nothing passes close to the singularity at $0$. – robjohn May 11 '19 at 02:25

I have a generalized elementary method for this problem,If f (x) is an even function, and the period is $\pi$,we have: $$\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx=\int_{0}^\frac{\pi}{2}f(x)g_n(x)\sin^nxdx \qquad (1)$$

Where the $g_n(x)$ in (1) is as follows $$g_n(x)=\begin{cases}\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\csc x\right),& \text{for n is odd $n\in\Bbb N$ and}\\[2ex] \frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\cot x\right),& \text{ for n is even .} \end{cases}$$ —————————————————————————————————————————————————— Proof: \begin{align} \int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x}\right)^ndx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin x}{x}\right)^ndx\\ &=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx\\ &=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin x}{x-k\pi}\right)^ndx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^n}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^n}+\frac{1}{(x-k\pi)^n}\right]\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_n(x)dx \end{align} We know by the Fourier series \begin{align} \csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\ \end{align} and \begin{align} \cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right) \end{align} Take the n-1 order derivative,thus we obtain $g_n(x)$. —————————————————————————————————————————————————— Example: \begin{align} (1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\ &=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\ &=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\ &=\frac{\pi}{4}\\ \end{align} \begin{align} (2.) \int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\ \end{align} \begin{align} (3.) \int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\ (4.) \int_{0}^{\infty}\frac{1}{3+\cos2x}\frac{\sin^2x}{x^2}dx &=\int_{0}^{\frac{\pi}{2}}\frac{1}{3+\cos2x}dx =\frac{\pi}{4\sqrt{2}}\\ \end{align}

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    A different integral that is related to this:https://math.stackexchange.com/questions/2406462/how-to-find-the-closed-form-of-the-integral-int-0-inftyfx-frac-sinnx – JamesJ Aug 28 '17 at 08:07

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).

A Blumenthal
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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
  • @robjohn Yeah, I suppressed the part of the argument which uses the fact that $n$ is odd; it comes up in the 'reorganization' of the terms from all these integrals. The even argument is similar but requires different bookkeeping. – A Blumenthal Feb 19 '13 at 20:33

There is an excellent result related with this integral, enjoy !

$$I=\int_{0}^{\infty }x^{p}\ \left ( \frac{\sin(x)}{x} \right )^ndx\quad, n=1,2,3...... , \quad 0\geq p\geq -1\\ \\ \\ I=\frac{\pi }{2(2i)^{n}\Gamma (n-p)}\sum_{m=0}^{n }(-1)^{n-m}\frac{n!}{m!(n-m)!}\left | n-2m \right |^{n-p-1}\left ( \frac{1}{\sin(\frac{n-p+1}{2})\pi }-\frac{\text{sgn}(n-2m)}{\sin(\frac{n-p}{2})\pi}i \right )$$

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Ahmed Hejazi
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