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?
I have a way of showing this directly, using Laplace transforms. In case you are new to this, a Laplace transform of a function $f(t)$ is
$$\hat{f}(s) = \int_0^{\infty} dt \, f(t) \, e^{-s t}$$
We begin with the following interesting formula. If $\hat{f}$ and $\hat{g}$ are the Laplace transforms of $f$ and $g$, respectively, then
$$\int_0^{\infty} du \, f(u) \hat{g}(u) = \int_0^{\infty} du \, \hat{f}(u) g(u)$$
Here, let $f(t) = \sin^{2 n+1}(t)$ and $\hat{g}(s) = 1/s$. Hopefully, it is clear that $g(t) = 1$. The computation of $\hat{f}(s)$, however, is a bit messy, albeit possible:
$$\begin{align} \hat{f}(s) &= \int_0^{\infty} dt \, \sin^{2 n+1}{(t)} \, e^{-s t} \\ &= \frac{1}{(2 i)^{2 n+1}} \sum_{k=0}^{2 n+1} (-1)^k \binom{2 n+1}{k} \int_0^{\infty} dt \, e^{i(2 n+1-2 k)t} e^{-s t}\\&=\frac{1}{(2 i)^{2 n+1}} \sum_{k=0}^{2 n+1} (-1)^k \binom{2 n+1}{k} \frac{1}{s-i(2 n+1-2 k)}\\&= \frac{(-1)^n}{2^{2 n}} \sum_{k=0}^n (-1)^k \binom{2 n+1}{k} \frac{2 n+1-2 k}{s^2+(2 n+1-2 k)^2}\end{align}$$
For that last step, I combined the $k$th and $(n-k)$th terms in the previous sum. The quantity we seek is
$$\begin{align}\int_0^{\infty} dx \frac{\sin^{2 n+1}{x}}{x} &= \frac{(-1)^n}{2^{2 n}} \sum_{k=0}^n (-1)^k \binom{2 n+1}{k} \int_0^{\infty} du \frac{2 n+1-2 k}{u^2+(2 n+1-2 k)^2}\\&= \frac{(-1)^n \pi}{2^{2 n+1}} \sum_{k=0}^n (-1)^k \binom{2 n+1}{k} \end{align}$$
It turns out that
$$\sum_{k=0}^n (-1)^k \binom{2 n+1}{k} = (-1)^n \binom{2 n}{n}$$
which fortuitously leads us to the stated result:
$$\int_0^{\infty} dx \frac{\sin^{2 n+1}{x}}{x} = \frac{\pi}{2^{2 n+1}} \binom{2 n}{n}$$
ADDENDUM
For a proof of the above sum, we use Knuth's negative index trick:
$$(-1)^k \binom{r}{k} = \binom{k-r-1}{k}$$
along with the well-known analog of a definite integral of a monomial:
$$\sum_{k=0}^n \binom{k+m}{k} = \binom{n+m+1}{n}$$
so that we have
$$\begin{align}\sum_{k=0}^n (-1)^k \binom{2 n+1}{k} &= \sum_{k=0}^n \binom{k-2 n-2}{k}\\&= \binom{n-2 n-1}{n} \\ &= \binom{-(n+1)}{n} \\ &= (-1)^n\binom{n+n+1-1}{n} \\ &= (-1)^n \binom{2 n}{n}\end{align}$$
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.
