Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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.

share|cite|improve this question
Have you tried it for small $n$, like $n=0$ and $n=1$? – Thomas Andrews Jul 17 '12 at 18:56
@Arjang: Please try to space your edits. The front page looks like a lot of the same question. – Asaf Karagila Mar 1 '13 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 '13 at 22:34

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 $$ \sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n} $$

share|cite|improve this answer
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: $$ \begin{align} \int_0^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x} &=\frac12\int_{-\infty}^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\int_{-\infty}^\infty\left(e^{ix}-e^{-ix}\right)^{2n+1}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\int_{\gamma^+}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &+\left(-\frac14\right)^{n+1}i\sum_{k=n+1}^{2n+1}(-1)^k\binom{2n+1}{k}\int_{\gamma^-}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}2\pi i\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}(-1)^n\binom{2n}{n}\\ &=\frac{1}{4^n}\frac{\pi}{2}\binom{2n}{n} \end{align} $$

share|cite|improve this answer
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

One more just for luck...

Use the evenness of the integrand, the binomial expansion of $\sin(x)^{2n}$ in terms of exponentials, and the Fourier transform representation of the rectangular function and you have:

\begin{aligned} \frac{1}{2}\int _{-\infty}^{\infty }\!{\frac { \sin \left( x \right) ^{ 2\,n+1}}{x}}{dx}&=\frac{1}{{2}^{2n+1}}\sum _{k=0}^{2\,n} {2\,n\choose k} \left( -1 \right) ^{n-k}\int _{-\infty }^{\infty }\!{\frac {\sin \left( x \right) {{\rm e}^{-2ix \left( n-k \right) }}}{x}}{dx}\\ &=\frac {\pi }{{2}^{2n+1}}\sum _{k=0}^{2\,n}{2\,n\choose k} \left( -1 \right) ^{n-k} \cases{1 &$ \left| n-k \right| <1/2$\cr 1/2 &$ \left| n-k \right| =1/2$\cr 0&$ \left| n-k \right|>1/2 $\cr}\\ &=\frac{\pi}{{2}^{2n+1}}{2\,n\choose n} \end{aligned} The rectangular function advantageously shows us that the only non-zero-weighted term in the sum is the $k=n$ term and we are spared any further manipulation or evaluation of sums.

share|cite|improve this answer
Thanks Graham. Ol' Fourier comes in handy :) – Cody Aug 9 '13 at 17:40

There is a theorem that states if $f(x)$ is continuous and $\pi$-peridodic on $\mathbb{R}$, then $$ \displaystyle\int_{-\infty}^{\infty} \frac{\sin x}{x} f(x) \ dx = \int_{0}^{\pi} f(x) \ dx. $$

See Graham Hesketh's comment for a way to prove this.

Using this theorem, $$ \begin{align} \int_{0}^{\infty} \frac{\sin^{2n+1} (x)}{x} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2n+1} (x)}{x} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin x}{x} \sin^{2n} (x) \ dx \\ &= \frac{1}{2} \int_{0}^{\pi} \sin^{2n} (x) \ dx = \int_{0}^{\frac{\pi}{2}} \sin^{2n} (x) \ dx \\ &= \frac{\pi}{2^{2n+1}} \binom{2n}{n}. \tag{1} \end{align}$$


share|cite|improve this answer
(+1) To prove that theorem write the integral as the inverse Fourier transform of a rect function, FT(sinc), multiplied by a weighted delta comb, FT(f(x)), (having applied the convolution theorem) and show that the only non-zero weighted delta term is the n=0 term which has the integral representation you give on the right. – Graham Hesketh Aug 10 '13 at 23:59

I am just adding the proof of the identity for those who have interest: $$ \sin^{2n+1} x = \frac{1}{4^n}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$ Using the complex representation and the Binomial Theorem, we have $$\begin{aligned} \sin^{2n+1}x&=\left(\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}\right)^{2n+1}\\ &=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}\binom{2n+1}{k}\mathrm{e}^{i(2n+1-k)x}(-1)^k\mathrm{e}^{i(-kx)}\\ &=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\mathrm{e}^{i(2(n-k)+1)x}\\ &=\frac{(-1)^n}{2^{2n+1}i}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\left[\cos\left(\left(2(n-k)+1\right)x\right) + i\sin\left(\left(2(n-k)+1\right)x\right)\right]\\ &=\frac{(-1)^n}{2^{2n+1}}\sum_{k=0}^{2n+1}(-1)^k\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right] \end{aligned} $$

Now, observe that $$\begin{aligned} \sum_{k=0}^{2n+1} a_{k} &= \sum_{k=0}^{n}a_{k}+\sum_{k=n+1}^{n+n+1}a_{k}\\ &=\sum_{k=0}^{n}a_{k}+\sum_{k=0}^{n}a_{n+1+k}\\ &=\sum_{k=0}^{n}a_{k}+\sum_{k=0}^{n}a_{n+1+n-k}\\ &=\sum_{k=0}^{n}\left(a_{k}+a_{2n+1-k}\right) \end{aligned} $$ Apply with $a_{k}=(-1)^{k}\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]$, so $$\begin{aligned} a_{2n+1-k}&=-(-1)^{k}\binom{2n+1}{2n+1-k}\left[-\sin\left(\left(2(n-k)+1\right)x\right) - i\cos\left(\left(2(n-k)+1\right)x\right)\right]\\ &=(-1)^{k}\binom{2n+1}{k}\left[\sin\left(\left(2(n-k)+1\right)x\right) + i\cos\left(\left(2(n-k)+1\right)x\right)\right]. \end{aligned} $$ Then, $$ a_{k}+a_{2n+1-k}=2(-1)^{k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$ Therefore, $$\begin{aligned} \sin^{2n+1} x&=\frac{1}{4^{n}}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right)\\ &=\frac{1}{4^n}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{n-k}\sin\left((2k+1)x\right)\\ &=\frac{1}{4^n}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{n+k+1}\sin\left((2k+1)x\right), \end{aligned}$$ as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.