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.

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    $\begingroup$ Have you tried it for small $n$, like $n=0$ and $n=1$? $\endgroup$ – Thomas Andrews Jul 17 '12 at 18:56
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    $\begingroup$ @Arjang: Please try to space your edits. The front page looks like a lot of the same question. $\endgroup$ – Asaf Karagila Mar 1 '13 at 22:33
  • $\begingroup$ @AsafKaragila : Thank you, by space do you mean time wise? I was just thinking I should wait between edits, or something else? $\endgroup$ – 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 $$ \sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n} $$

  • $\begingroup$ Wht did you do to get from the first line to the second? (after "We get") $\endgroup$ – Dennis Gulko Jul 17 '12 at 19:27
  • $\begingroup$ This was a simple change of variables, for $c>0$, $\int_0^\infty \frac{\sin(c x)}{x} \mathrm{d} x = \int_0^\infty \frac{\sin(c x)}{c x} \mathrm{d} (c x) \stackrel{y=cx}{=} \int_0^\infty \frac{\sin(y)}{y} \mathrm{d} y$. $\endgroup$ – Sasha Jul 17 '12 at 19:33
  • $\begingroup$ Yeah, sorry. I figured that out, but couldn't delete my comment (from my phone) :-) $\endgroup$ – Dennis Gulko Jul 17 '12 at 19:43
  • $\begingroup$ Wow, thanks Sasha. Very nice and elegant. $\endgroup$ – Cody Jul 17 '12 at 20:41
  • $\begingroup$ 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) $\endgroup$ – 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: $$ \begin{align} \int_0^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x} &=\frac12\int_{-\infty}^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\int_{-\infty}^\infty\left(e^{ix}-e^{-ix}\right)^{2n+1}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\int_{\gamma^+}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &+\left(-\frac14\right)^{n+1}i\sum_{k=n+1}^{2n+1}(-1)^k\binom{2n+1}{k}\int_{\gamma^-}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\ &=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}2\pi i\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\\ &=\left(-\frac14\right)^{n}\frac{\pi}{2}(-1)^n\binom{2n}{n}\\ &=\frac{1}{4^n}\frac{\pi}{2}\binom{2n}{n} \end{align} $$

  • $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ Thanks Graham. Ol' Fourier comes in handy :) $\endgroup$ – 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, $$ \begin{align} \int_{0}^{\infty} \frac{\sin^{2n+1} (x)}{x} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2n+1} (x)}{x} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin x}{x} \sin^{2n} (x) \ dx \\ &= \frac{1}{2} \int_{0}^{\pi} \sin^{2n} (x) \ dx = \int_{0}^{\frac{\pi}{2}} \sin^{2n} (x) \ dx \\ &= \frac{\pi}{2^{2n+1}} \binom{2n}{n}. \tag{1} \end{align}$$

$(1)$ http://en.wikipedia.org/wiki/Wallis%27_integrals

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    $\begingroup$ (+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. $\endgroup$ – Graham Hesketh Aug 10 '13 at 23:59
  • $\begingroup$ This is the famous Lobachevsky integral theorem $\endgroup$ – FofX Sep 5 '18 at 16:46

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.


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