$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ 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.
 A: There is a theorem that states if $f(x)$ is continuous and $\pi$-periodic 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
A: 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.
A: 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
$$
  \sum_{k=0}^n (-1)^k \binom{2n+1}{n+k+1} = \sum_{k=0}^n \left(g(k+1)-g(k)\right) = g(n+1) - g(0) = -g(0) = \binom{2n}{n}
$$
A: 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.
A: 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:
$$
\begin{align}
\int_0^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x}
&=\frac12\int_{-\infty}^\infty\sin^{2n+1}(x)\frac{\mathrm{d}x}{x}\\
&=\left(-\frac14\right)^{n+1}i\int_{-\infty}^\infty\left(e^{ix}-e^{-ix}\right)^{2n+1}\frac{\mathrm{d}x}{x}\\
&=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\int_{\gamma^+}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\
&+\left(-\frac14\right)^{n+1}i\sum_{k=n+1}^{2n+1}(-1)^k\binom{2n+1}{k}\int_{\gamma^-}e^{ix(2n-2k+1)}\frac{\mathrm{d}x}{x}\\
&=\left(-\frac14\right)^{n+1}i\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}2\pi i\\
&=\left(-\frac14\right)^{n}\frac{\pi}{2}\sum_{k=0}^{n}(-1)^k\binom{2n+1}{k}\\
&=\left(-\frac14\right)^{n}\frac{\pi}{2}(-1)^n\binom{2n}{n}\\
&=\frac{1}{4^n}\frac{\pi}{2}\binom{2n}{n}
\end{align}
$$
