# $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral

Here is a fun integral I am trying to evaluate:

$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$

I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.

But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach.

$$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$

or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.

Thanks everyone.

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1
Have you tried it for small $n$, like $n=0$ and $n=1$? – Thomas Andrews Jul 17 '12 at 18:56

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@Arjang: Please try to space your edits. The front page looks like a lot of the same question. – Asaf Karagila Mar 1 at 22:33
@AsafKaragila : Thank you, by space do you mean time wise? I was just thinking I should wait between edits, or something else? – Arjang Mar 1 at 22:34

## 2 Answers

Using $$\sin^{2n+1}(x) = \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} \sin\left((2k+1)x\right)$$ We get $$\begin{eqnarray} \int_0^\infty \frac{\sin^{2n+1}(x)}{x}\mathrm{d} x &=& \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left((2k+1)x\right)}{x}\mathrm{d} x\\ &=& \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left(x\right)}{x}\mathrm{d} x \\ &=& \frac{\pi}{2^{2n+1}}\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \frac{\pi}{2^{2n+1}} \binom{2n}{n} \end{eqnarray}$$ The latter sum is evaluated using telescoping trick: $$\sum_k (-1)^k \binom{2n+1}{n+k+1} = \sum_k (-1)^k \frac{2n+1}{n+k+1} \binom{2n}{n+k} = (-1)^{k+1} \binom{2n}{n+k} =: g(k)$$ meaning that $$g(k+1) - g(k) = (-1)^k \binom{2n+1}{n+k+1}$$ Hence $$\sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n}$$

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Wht did you do to get from the first line to the second? (after "We get") – Dennis Gulko Jul 17 '12 at 19:27
This was a simple change of variables, for $c>0$, $\int_0^\infty \frac{\sin(c x)}{x} \mathrm{d} x = \int_0^\infty \frac{\sin(c x)}{c x} \mathrm{d} (c x) \stackrel{y=cx}{=} \int_0^\infty \frac{\sin(y)}{y} \mathrm{d} y$. – Sasha Jul 17 '12 at 19:33
Yeah, sorry. I figured that out, but couldn't delete my comment (from my phone) :-) – Dennis Gulko Jul 17 '12 at 19:43
Wow, thanks Sasha. Very nice and elegant. – Cody Jul 17 '12 at 20:41
My method is essentially the same, but steps have been rearranged. With yours, you get to use $\large\int_0^\infty\frac{\sin(x)}{x}\,\mathrm{d}x$. (+1) – robjohn♦ Jul 17 '12 at 23:38

Since $\dfrac{\sin^{2n+1}(x)}{x}$ is an even function, we can integrate over the whole real line and divide by $2$.

Write $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$. Since there are no singularities and the integrand vanishes as $|x|\to\infty$, we can move the path of integration in the direction of $-i$. Expand using the binomial theorem, and close the paths of integration in two ways: for the integrands with $e^{+ikx}$ circle back counter-clockwise around the upper half-plane ($\gamma^+$); for the integrands with $e^{-ikx}$ circle back clockwise around the lower half-plane ($\gamma^-$).

Note that $\gamma^-$ contains no poles, so those integrals can be ignored.

We will use the identity $$\begin{align} \sum_{k=0}^m(-1)^k\binom{n}{k} &=\sum_{k=0}^m(-1)^k\binom{n}{k}\binom{m-k}{m-k}\\ &=(-1)^m\sum_{k=0}^m\binom{n}{k}\binom{-1}{m-k}\\ &=(-1)^m\binom{n-1}{m} \end{align}$$ Finally, to the point: $$\begin{align} \int_0^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x} &=\frac12\int_{-\infty}^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\int_{-\infty}^\infty\left(e^{ix}-e^{-ix}\right)^{2n+1}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\int_{\gamma^+}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &+\left(-\frac14\right)^{n+1}i\sum_{k=n+1}^{2n+1}(-1)^k\binom{2n+1}{k}\int_{\gamma^-}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}2\pi i\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}(-1)^n\binom{2n}{n}\\ &=\frac{1}{4^n}\frac{\pi}{2}\binom{2n}{n} \end{align}$$

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There it is RobJohn!!!. :):) That is along the lines I was thinking, but I got discombobulated in all of that. Thanks much.  Your use of contours was clever. – Cody Jul 17 '12 at 23:46
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