Sine all the Way

Calculus Level 4

$\large I_n = \int_0^{\pi/2}{\dfrac{\sin((2n + 1)x)}{\sin(x)} dx}$

For $I_n$ as defined above, find the value of $\displaystyle \lim_{n \to \infty}{2I_n}$ .

1 solution

Rohan Shinde
May 4, 2018

Using the substitution $u=2x$ and $du=2dx$

The integral changes to $\frac 12\int_0^{\pi} \frac {\sin\left(\left(n+\frac 12\right)u\right)}{\sin \frac {u}{2}}du$

And inside the integral is just the Dirichlet kernel i.e. $D_n$

Hence $I_n=\frac 12 \left(\int_0^{\pi} 1+2\sum_{k=1}^n \cos (ku) du\right)$

Which simplifies to $I_n=\frac {\pi}{2} +\sum_{k=1}^n \int_0^{\pi} \cos (ku) du$

The integral along with summation simply becomes $0$ .

Hence $\lim_{n\to \infty} 2I_n=\lim_{n\to \infty} 2\cdot \frac {\pi}{2}=\pi$