Integral of the day XIX

Integral involving a finite series

Today we show the proof of the following integral posted by @infseriesbot \[\int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx = \frac{1}{2^{n+1}}\sum_{k=1}^{n} \frac{2^k}{k}\] The proof will rely on the Euler's formula and some properties of the binomial coefficient.

Proof

\begin{align*} \int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx =& \Im\left( \int_{0}^{\frac{\pi}{2}} e^{inx}\cos^n(x) dx \right) \textrm{ (Euler's formula)} \\ =& \Im\left( \int_{0}^{\frac{\pi}{2}} e^{inx}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^ndx \right)\\ =& \Im\left( \frac{1}{2^n}\int_{0}^{\frac{\pi}{2}} e^{inx}\sum_{j=0}^{n}\binom{n}{j} e^{ix(n-2j)}dx \right) \quad \textrm{(Binomial expansion)}\\ =& \Im\left( \frac{1}{2^n}\sum_{j=0}^{n} \binom{n}{j}\int_{0}^{\frac{\pi}{2}} e^{i2x(n-j)}dx \right)\\ =&\Im\left( \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \left[\cos\left(2x(n-j)\right)+i\sin\left(2x(n-j)\right) \right]dx\right) \textrm{ (Euler's formula)}\\ =&\Im\left( \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \left[\int_{0}^{\frac{\pi}{2}} \cos\left(2x(n-j)\right) dx +i \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx \right]\right)\\ =&\Im\left( \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \cos\left(2x(n-j)\right) dx +i \frac{1}{2^n}\sum_{j=0}^{n} \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx \right)\\ =& \frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx \\ =& \frac{1}{2^n}\sum_{j=0}^{n-1}\binom{n}{j} \int_{0}^{\frac{\pi}{2}} \sin\left(2x(n-j)\right) dx + \underbrace{\binom{n}{n} \int_{0}^{\frac{\pi}{2}} \sin\left(0\right) dx}_{=0} \\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1}{n-j} \int_{0}^{\pi(n-j)} \sin(w) dw \quad \left( w \mapsto 2x(n-j) \right)\\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1}{n-j} \int_{0}^{\pi(n-j)} \sin(w) dw \\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1}{n-j} \left[-\cos(w)\right]_{0}^{\pi(n-j)}\\ =& \frac{1}{2^{n+1}}\sum_{j=0}^{n-1}\binom{n}{j}\frac{1-\cos\left(\pi(n-j)\right)}{n-j} \\ =& \frac{1}{2^{n+1}}\sum_{k=1}^{n}\binom{n}{n-k}\frac{1-\cos\left(\pi k \right)}{k} \\ =& \frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{n-k}\frac{\sin^2\left(\frac{\pi k}{2} \right)}{k} \quad \left(\sin^2 \theta =\frac{1-\cos(2\theta)}{2}\right)\\ =& \frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{k}\frac{\sin^2\left(\frac{\pi k}{2} \right)}{k} \;\;\quad \binom{n}{n-m} = \binom{n}{m} \\ \end{align*} Note that \[ \left(\sin^2\left(\frac{\pi k}{2} \right)\right)_{k\in \mathbb{N}} = (1,0,1,0,1,0,...)\] Hence \begin{align*} I = \int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx =& \frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{k}\frac{\sin^2\left(\frac{\pi k}{2} \right)}{k}\\ =\frac{1}{2^{n}}\sum_{k=1}^{n}\binom{n}{2k-1}\frac{1}{2k-1}\\ \end{align*} Finally, we are going to prove that \[ 2\sum_{k=1}^{n}\binom{n}{2k-1}\frac{1}{2k-1} = \sum_{k=1}^{n} \frac{2^k}{k} \quad \forall n\in \mathbb{N} \] using mathematical induction:

Base case $n=1$ \[2\binom{1}{2-1}\frac{1}{2-1} = 2 \] \[ \frac{2^1}{1} = 2 \] Inductive step, suppose this is valid for $n=m$ \[ 2\sum_{k=1}^{m}\binom{m}{2k-1}\frac{1}{2k-1} = \sum_{k=1}^{m} \frac{2^k}{k} \] We need to prove this for $\underline{n= m+1}$

From the recursion formula \[ \binom{v+1}{m} =\binom{v}{m} + \binom{v}{m-1} \tag{1} \] and the expression \[ \sum_{j=0}^{J} \binom{n}{j} = 2^n \quad J\geq n \tag{2}\] we have \begin{align*} 2\sum_{k=1}^{m+1}\binom{m+1}{2k-1}\frac{1}{2k-1} = & 2\sum_{k=1}^{m}\binom{m+1}{2k-1}\frac{1}{2k-1} + \underbrace{2\binom{m+1}{2m+1}\frac{1}{2m+1}}_{=0}\\ =& 2\sum_{k=1}^{m}\left[\binom{m}{2k-1}+\binom{m}{2k-2}\right]\frac{1}{2k-1} \quad \textrm{ from (1)}\\ =& 2\sum_{k=1}^{m}\binom{m}{2k-1}\frac{1}{2k-1}+2\sum_{k=1}^{m}\binom{m}{2k-2}\frac{1}{2k-1}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+2\sum_{k=1}^{m}\binom{m}{2k-2}\frac{1}{2k-1} \quad \textrm{(inductive hypothesis)}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+2\sum_{j=0}^{m-1}\binom{m}{2j}\frac{1}{2j+1} \quad (j = k-1)\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+2\sum_{j=0}^{m-1}\frac{m!}{(m-2j)!(2j)!(2j+1)} \\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{m-1}\frac{(m+1)!}{(m-2j)!(2j+1)!} \\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{m-1}\binom{m+1}{2j+1} \\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{m-1}\left[\binom{m}{2j+1}+ \binom{m}{2j}\right] \quad \textrm{ from (1)}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2}{m+1}\sum_{j=0}^{2m-1}\binom{m}{j}\\ =& \sum_{k=1}^{m} \frac{2^k}{k}+\frac{2^{m+1}}{m+1}\quad \textrm{ from (2)}\\ =& \sum_{k=1}^{m+1} \frac{2^k}{k} \end{align*} Hence, we can conclude \[\boxed{\int_{0}^\frac{\pi}{2} \sin(nx)\cos^n(x) dx = \frac{1}{2^{n+1}}\sum_{k=1}^{n} \frac{2^k}{k}}\]