Inegral of the day III

Integral involving the binomial coefficient

Today we show the proof of the following integral involving the binomial coefficient posted by @infseriesbot. The bot has a typo but this is the correct formula: \[ \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx = \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2}\]
Proof

We need the following ingredients for the proof:


1. $\sin^{r}(x)$ can be expanded with the following formula:
\[ sin^{r}(x) = \begin{cases} \displaystyle \frac{(-1)^n}{2^{r-1}} \sum_{j=0}^{n-1} (-1)^j \binom{r}{j} \cos\left[(r-2j)x\right] + \frac{1}{2^n} \binom{r}{n} &\quad \textrm{if} \quad r=2n \\ \displaystyle \frac{(-1)^n}{2^{r-1}}\sum_{j=0}^{n} (-1)^j \binom{r}{j} \sin\left[(r-2j)x\right] & \quad \textrm{if} \quad r=2n+1 \end{cases} \]

Proof

\begin{align*} (2i\sin(x))^{r} = (e^{ix}-e^{-ix})^{r} = & \sum_{j=0}^{r} \binom{r}{j} (-1)^{r-j}e^{ixj}e^{-ix(r-j)} =\sum_{j=0}^{r} \binom{r}{j} (-1)^{r-j}e^{-ix(r-2j)} \\ =& \sum_{j=0}^{r} \binom{r}{j}(-1)^{r-j}\left[\cos(x(r-2j))-i\sin(x(r-2j))\right] \end{align*} Now suppose that $r=2n+1$, then \begin{align*} 2^{2n}(-1)^{n}(2i)\sin^{2n+1}(x) = &\sum_{j=0}^{2n+1} \binom{2n+1}{j}(-1)^{2n+1-j} \left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\ =& \sum_{j=0}^{n} \binom{2n+1}{j}(-1)^{2n+1-j} \left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\ +& \sum_{j=n+1}^{2n+1} \binom{2n+1}{j}(-1)^{2n+1-j}\left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\ =& \sum_{j=0}^{n} \binom{2n+1}{j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]\\ +&\sum_{j=n+1}^{2n+1} \binom{2n+1}{2n+1-j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right] \end{align*} Lets $k=2n+1-j$ then \begin{align*} &\sum_{j=n+1}^{2n+1} \binom{2n+1}{2n+1-j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]\\ =& \sum_{k=0}^{n} \binom{2n+1}{k}(-1)^{k}\left[\cos(-x(2n+1-2k))-i\sin(-x(2n+1-2k))\right] \end{align*} Given that for $k=0,...,n$ \begin{eqnarray*} \cos(-x(2n+1-2k))-\cos(x(2n+1-2k))&=&0\\ i\sin(x(2n+1-2k))-i\sin(-x(2n+1-2k))& =& 2i\sin(x(2n+1-2k)) \end{eqnarray*} \begin{equation} \therefore sin^{2n+1}(x) =\frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \sin\left[(2n+1-2j)x\right] \end{equation} A similar argument shows the result for $r=2n$


2. From complex analysis we know that (proof in the appendix):


\begin{equation} \int_{0}^{\infty} \frac{\sin(x)}{x} = \frac{\pi}{2} \end{equation}


3. From combinatorics we kwnow that that (proof in the appendix):


\begin{equation} \sum _{j=0}^{k} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{k} \end{equation} We just have tu put all the ingredients together \begin{align*} \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx =& \int_{0}^{\infty} \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx \quad \textrm{from (1)}\\ =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx \end{align*} If $u= (2n+1-2j)x $ then $ \displaystyle dx = \frac{1}{2n+1-2j} du$ \begin{align*} \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx\\ =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin(u)}{u} du\\ \overset{(2)}{=}& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \frac{\pi}{2} \quad \textrm{from (2)}\\ =& \frac{(-1)^n}{2^{2n}} (-1)^{n} \binom{2n}{n}\frac{\pi}{2} \quad \textrm{from (3)}\\ =& \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2} \end{align*} Therefore \[\boxed{ \int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx = \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2}}\]

Appendix: Here are the proofs of (2) and (3):

Proof of (2)

\[\int_{0}^{\infty} \frac{\sin(x)}{x} = \frac{\pi}{2} \] The integral \[I = \operatorname{p.v.} \int_{-\infty}^{\infty} \frac{e^{ix}}{x}dx \] exists by the \emph{Cauchy principal value theroem}. Therefore \[\operatorname{Im}I = \operatorname{p.v.} \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx\] also exists. Then \[\operatorname{p.v.} \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx = \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx \] by the continuity of $\displaystyle \frac{\sin(x)}{x}$ at $x=0$. The existence of $\displaystyle \int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx $ also implies the existence of $\displaystyle \int_{0}^{\infty} \frac{\sin(x)}{x}dx $ and \[\operatorname{Im} I = 2\int_{0}^{\infty} \frac{\sin(x)}{x}dx\] Now applying the residue formula for the principal value: \[ I = \pi i \sum \underset{z=0}{\textrm{Res}}\left(\frac{e^{ix}}{x}\right)=\pi i \Longrightarrow \int_{0}^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}\]

Proof of (3)

If $k,m \in \mathbb{N}$ \[ \sum _{j=0}^{k} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{k}\] Base case Lets $k=1$ then \[(-1)^{0}\binom{m}{0} +(-1)^{1}\binom{m}{1} = 1-m\] On the other hand \[(-1)^{1}\binom{m-1}{1} = 1-m\] Inductive hypothesis Assume the proposition is true for $k=n$, then \[\sum _{j=0}^{n} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{n}\] Inductive step Lets $k=n+1$ then \begin{align*} \sum _{j=0}^{n+1} (-1)^j\binom{m}{j} = &\sum _{j=0}^{n} (-1)^j\binom{m}{j} + (-1)^{n+1}\binom{m}{m+1}\\ =& (-1)^{n} \binom{m-1}{n} +(-1)^{n+1} \left[\binom{m-1}{n} +\binom{m-1}{n+1}\right]\\ =& (-1)^{n+1}\binom{m-1}{n+1} \end{align*}