TITLE: Proving that $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}$ by induction
QUESTION [14 upvotes]: I need to prove
$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$
I've seen other demonstrations of this, but they use some identities that I don't understand. So I asked my calculus teacher and he told me I can prove that by induction. I've already done the case $n=1$, but I can do the rest, could you help me please?

REPLY [6 votes]: Here's another way to do it (though it doesn't use induction). Write $\sin{x} = (e^{ix} - e^{-ix})/2i$. Then
$$
(\sin{x})^{2n+1} = {1\over(2i)^{2n+1}}\sum_{k = 0}^{2n+1}(-1)^k{2n+1\choose k} e^{i(2n+1-2k)x}.
$$
Now, since your integrand is even, we can write
$$
\begin{align}
2\int_0^\infty {(\sin{x})^{2n+1}\over x}\,dx & = \int_{-\infty}^\infty {(\sin{x})^{2n+1}\over x}\,dx \\
& = \lim_{\delta\to 0}\int_{|x|>\delta} {(\sin{x})^{2n+1}\over x}\,dx \\
& = {1\over(2i)^{2n+1}}\sum_{k = 0}^{2n+1}(-1)^k{2n+1\choose k} \lim_{\delta\to 0}\int_{|x|>\delta} {e^{i(2n+1-2k)x}\over x}\,dx.
\end{align}
$$
The real part of $e^{i(2n+1-2k)x}/x$ is $\cos\{(2n+1-2k)x\}/x$, which is an odd function of $x$; hence the integral of this real part over $|x| > \delta$ vanishes for each fixed $R$. So in fact 
$$
\begin{align}
\lim_{\delta\to 0}\int_{|x|>\delta} {e^{i(2n+1-2k)x}\over x}\,dx &= i\lim_{\delta\to 0}\int_{|x|>\delta} {\sin\{(2n+1-2k)x\}\over x}\,dx \\
& = i\int_{-\infty}^\infty {\sin\{(2n+1-2k)x\}\over x}\,dx.
\end{align}
$$
A simple substitution shows that (using the value of your integral when $n = 0$)
$$
\begin{align}
\int_{-\infty}^\infty {\sin\{(2n+1-2k)x\}\over x}\,dx & = \pi\operatorname{sign}(2n+1-2k) = \left\{\begin{array}{ll}
\pi & 0\leq k\leq n\\
-\pi & k\geq n+1.
\end{array}\right.
\end{align}
$$
Finally, using the fact that $$(-1)^k{2n+1\choose k} = -(-1)^{2n+1-k}{2n+1\choose 2n+1-k},$$ we get
$$
\begin{align}
2\int_0^\infty {(\sin{x})^{2n+1}\over x}\,dx & = {1\over(2i)^{2n+1}}\sum_{k = 0}^{2n+1}(-1)^k{2n+1\choose k} i\int_{-\infty}^\infty {\sin\{(2n+1-2k)x\}\over x}\,dx \\
& = {i\pi\over(2i)^{2n+1}}\left(\sum_{k = 0}^{n} - \sum_{k = n+1}^{2n+1}(-1)^k{2n+1\choose k}\right) \\
& = {2\pi(-1)^n\over2^{2n+1}}\sum_{k = 0}^{n}(-1)^k{2n+1\choose k}.
\end{align}
$$
Ron Gordon has already proved that the sum in this expression is equal to $(-1)^n{2n\choose n}$, so I'll leave it at that.