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Exercise. Evaluate the improper trigonometric integral $$\int_{-\infty}^{\infty}\frac{\sin^n(x)}{x^n}\mathrm{d}x, \ \ \ n\in\Bbb{N}^+,$$ using the complex epsilon method.

Let us view only odd powers, that is

$$\int_{-\infty}^{\infty}\frac{\sin^{2n+1}(x)}{x^{2n+1}}\mathrm{d}x.$$

I have shown that

$$\sin^{2n+1}(x) = \frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sin[(2n+1-2k)x]$$

which leaves us with evaluating

$$\lim\limits_{\varepsilon\to0^+}\int_{-\infty}^{\infty}\Im\overbrace{\left\{\frac{1}{z^{2n+1}+\varepsilon^{2n+1}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\exp[\mathrm{i}(2n+1-2k)z]\right\}}^{f(z)}\mathrm{d}z.$$

We have a total of $n+1$ simple poles of interest, the first $n$ lay in the upper half of the complex plane while the $(n+1)$th is on the real axis.

$$z_m = \varepsilon\exp\left[\mathrm{i}\frac{(2m+1)\pi}{2n+1}\right], \ \ \ m\in\{0,\ldots,n-1\}.$$

Thus

$$\int_{-\infty}^{\infty}\frac{\sin^{2n+1}(x)}{x^{2n+1}}\mathrm{d}x = \lim\limits_{\varepsilon\to0^+}\Im\left[2\pi \mathrm{i}\sum_{m=0}^{n-1}\operatorname{\underset{z=z_m}{Res}}f(z) + \pi\mathrm{i}\operatorname{\underset{z=z_n}{Res}}f(z)+ \lim\limits_{R\to\infty}\int_{C_R}f(z)\right].\tag1$$

From $(1)$ (since simple poles only)

$$\operatorname{\underset{z=z_m}{Res}}f(z)=\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\exp[\mathrm{i}(2n+1-2k)z_m]$$ or, equivalently,

$$\operatorname{\underset{z=z_m}{Res}}f(z)=\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sum_{j=0}^{\infty}\frac{[\mathrm{i}(2n+1-2k)z_m]^j}{j!}.$$

The terms of the infinite sum above indice $2n$ become irrelevant as $\varepsilon \to 0^+$. Therefore,

$$\operatorname{\underset{z=z_m}{Res}}f(z)\equiv\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sum_{j=0}^{2n}\frac{[\mathrm{i}(2n+1-2k)z_m]^j}{j!}.\tag2$$

  • TL; DR: What is next? How can one simplify the expression

$$\sum_{m=0}^{n-1}\frac{1}{(2n+1)(z_m)^{2n}}\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\\k\end{pmatrix}\sum_{j=0}^{2n}\frac{[\mathrm{i}(2n+1-2k)z_m]^j}{j!}?\tag3$$

EDIT: I have now accepted an answer that gets around this sum but yet still uses a similar method. However, note that simplifying $(3)$ as a sum remains an open problem for me, so an answer along that line would be entitled to a bounty (+50).

Linear Christmas
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  • Borwein & Borwein studied some problems related to your question. See this link. In particular, they have an "explicit" expression for your integral in page 6 $\left(~\mbox{see}\ \left(ii\right)~\right)$ – Felix Marin Apr 24 '17 at 04:20
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    $$ \int_{-\infty}^{\infty}{\sin^{n}\left(,x,\right) \over x^{n}},\mathrm{d}x = \left[1 + {1 \over 2^{n - 2}}\sum_{r = 1}^{\left\lfloor,{n/2},\right\rfloor}{\left(-1\right)^{r}\left(,n - 2r,\right)^{n - 1} \over \left(,r-1,\right)!\left(,n - r,\right)!}\right]\pi,,\qquad n \geq 2 $$ – Felix Marin Apr 24 '17 at 04:35

2 Answers2

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Once you get the Fourier sine series of $\sin(x)^n$, you just need to apply integration by parts multiple times, in order to convert the original integral in a combination of integrals of the form $$ \int_{-\infty}^{+\infty}\frac{\sin(mx)}{x}\,dx = \pi.$$

Linear Christmas
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Jack D'Aurizio
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  • Yep. That'll work. This way I will at least have a general result to check against if the sum ever reveals itself :=). Thanks! (Should I have asked two seperate questions, one for the integral itself and one question for the sum I am having trouble with?) – Linear Christmas Apr 23 '17 at 16:33
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    @LinearChristmas: I do not think so, they are quite connected. However, I seem to recall such question was already asked on MSE, it is probably the case to dig the database a bit. – Jack D'Aurizio Apr 23 '17 at 16:35
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It is easily seen that for some $\alpha$ $$\left(\frac{\sin x}{x - i\alpha}\right)^{2n+1}=\frac{(-1)^{n}}{i 2^{2n+1}}\sum_{k=0}^{2 n +1}(-1)^k {{2 n +1}\choose {k}} \frac{\exp [i(2\overline{n-k} + 1)x]}{(x - i \alpha)^{2n+1}}.$$ The required integral is the limit $\left[\int_{-\infty}^{\infty}\left(\frac{\sin x}{x - i\alpha}\right)^{2n+1}dx\right]_{\alpha\rightarrow 0^+}$. For the contour closing in the upper half plane, only terms upto $k = n$ in the above sum contribute to the value of the integral. One may rewrite $\exp [i(2\overline{n-k} + 1)x]$ as $\exp [i(2\overline{n-k} + 1)(x - i \alpha)]\exp[-(2\overline{n-k} + 1)\alpha]$ and the residue of each sumand such as $$\frac{\exp [i(2\overline{n-k} + 1)(x - i \alpha)]}{(x - i \alpha)^{2n+1}}$$ is clearly $\frac{i^{2n}(2\overline{n -k}+1)^{2n}}{(2n)!}$. After taking the relevant limit the integral reduces to the following sum $$\int_{-\infty}^{\infty}\left(\frac{\sin x}{x}\right)^{2n+1}dx = \frac{\pi}{(2n)!}\sum_{k=0}^{n}(-1)^k{{2n + 1}\choose {k}} \left(n-k+1/2\right)^{2n}$$ which seems simpler than what you obtained.

EDIT: As noted by the OP, the result can be immediately generalized for any positive integer $n$ to $$\int_{-\infty}^{\infty}\left(\frac{\sin x}{x}\right)^n dx = \frac{\pi}{2^{n-1}(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k{{n}\choose {k}} \left(n-2k\right)^{n-1}.$$

vnd
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  • What do you mean by an integer that has a bar over it? I have never seen this kind of notation. – Linear Christmas Apr 23 '17 at 19:06
  • $\overline{n-k}$ is same as $(n-k)$ – vnd Apr 23 '17 at 19:08
  • I have gone through your answer, and it looks correct. Is this $$\frac{(-1)^{n}}{i 2^{2n+1}}\sum_{k=0}^{n}(-1)^k {{2 n +1}\choose {k}}\frac{\exp [i(2\overline{n-k} + 1)]}{(x - i \alpha)^{2n+1}}$$ effectively $f(z)$ in this answer, in that all that is left is to show $$\lim\limits_{R\to\infty}\int_{C_R}f(z) = 0?$$ – Linear Christmas Apr 23 '17 at 20:14
  • By $C_R$ I meant the "open" semi-circle. That is, the entire closed semi-circle contour $C$ is $C_R$ plus the real line from $-R$ to $R$. In other words, the "open" semi-circle should not contribute to the integral (probably by Jordan's lemma). Am I right? – Linear Christmas Apr 23 '17 at 20:29
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    yes. That is correct. – vnd Apr 23 '17 at 20:31
  • Thanks again for a beautiful answer! If you are interested, note that setting $2n + 1 \to n$ gives $$\frac{\pi}{2^{n-1}(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k{{n}\choose {k}} \left(n-2k\right)^{n-1}$$ which also happens to work for even powers. – Linear Christmas Apr 23 '17 at 20:38
  • I redid the maths today, and seem to pick up an extra $2$ along the way :-). I think it's because we mean different things by manipulating the original sum to have an index of $n$ instead of $2n + 1$. I wrote the sum as $$\frac{\sin^{2n+1}(x)}{(x-\mathrm{i}\alpha)^{2n+1}} = \frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\k\end{pmatrix}\sin[(2n+1-2k)x]\frac{1}{(x-\mathrm{i}\alpha)^{2n+1}},$$ and then as $$ \Im\frac{(-1)^n}{4^n}\sum_{k=0}^{n}(-1)^k\begin{pmatrix}2n+1\k\end{pmatrix}\exp[i(2n+1-2k)z]\frac{1}{(z-\mathrm{i}\alpha)^{2n+1}}.$$ Could you clarify what you had in mind? – Linear Christmas Apr 25 '17 at 09:10
  • Instead just use the fact that $$(\sin x)^{2n+1} = \frac{(\exp(ix) - \exp(-ix))^{2n+1}}{(2 i)^{2n+1}} = \frac{(-1)^n}{i2^{2n+1}}\sum_{k=0}^{2n+1}{{2n+1}\choose{k}}(\exp(ix))^{(2n + 1 - k)}(-\exp(-ix))^k$$ – vnd Apr 25 '17 at 11:35
  • I'm starting to feel stupid :). I still don't understand why this would mean the terms above index $n$ have no effect on the integral. For example, $k = 2n$ in your sum gives residue of $$\frac{\mathrm{i}^{2n}}{(2n)!}$$ which (to me) there is no getting around unless one lowers the index to begin with. And in turn, that feels natural by awknowledging it's actually another sine function (and then later turn it back to an exponent to be bounded on the upper half to apply Jordan's lemma). – Linear Christmas Apr 25 '17 at 14:23
  • This is an example of what I mean: picture (for case $n = 1$) – Linear Christmas Apr 25 '17 at 14:46
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    For $k > n$, Jordan's lemma can be applied only if you close the contour in the lower half plane. But such a contour does not include any poles ($\alpha > 0$). So the contributions are all 0 for such $k$. – vnd Apr 25 '17 at 14:47
  • You've been very helpful. This makes sense. So effectively I will have a contour in the upper half plane for $k \leq n$ ($k = n$ can be bounded by means other than Jordan's lemma), and another contour in the lower half plane for all $k > n$, and since there are no poles there the usual Cauchy-Goursat theorem applies. – Linear Christmas Apr 25 '17 at 15:05
  • Let us continue this discussion in chat. – Linear Christmas Apr 25 '17 at 15:16