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

If $$I_n=\int_0^{\frac {\pi}{2}} \frac {\sin ((2n+1)x)dx}{\sin x}$$ then evaluate $$\lim_{n\to \infty} 2I_n$$

My try: $$I_n=\int_0^{\frac {\pi}{2}} \frac {\sin ((2n+1)x)dx}{\sin x}$$ $$=\int_0^{\frac {\pi}{2}} \frac {(\sin 2nx\cos x+\cos 2nx\sin x)dx}{\sin x}$$ $$=\int_0^{\frac {\pi}{2}} \sin (2nx)\cot (x) dx+\int_0^{\frac {\pi}{2}} \cos (2nx) dx$$

$$=\left(\int_0^{\frac {\pi}{2}} \sin (2nx)\cot (x) dx\right)+ \frac {\sin n\pi}{2n}$$

In the first integral I tried to use integration by parts and could reach up-to $$\left(\int_0^{\frac {\pi}{2}} \sin (2nx)\cot x dx\right)=\sin 2nx\ln (\sin x) -2n\int_0^{\frac {\pi}{2}} \cos 2nx\ln (\sin x) dx$$

But now I am stuck up here. Any help would be very beneficial.

• Commented May 5, 2018 at 4:39
• @lab bhattacharjee Thanks for the links. I'll take a look at them and see if they could help. :-) Commented May 5, 2018 at 4:45
• IMHO the way via the Dirichlet kernel is the easiest, as you need only to integrate some cosines and a constant and you will get nice sum. Commented May 5, 2018 at 6:08

Apologies for this first part being rather unintuitive.

\begin{aligned} I_n-I_{n-1} &= \int_0^{\frac\pi 2} \frac{\sin(2nx+x)-\sin(2nx-x)}{\sin x}\,dx \\ &= \int_0^{\frac\pi 2} \frac{2 \cos2nx \sin x}{\sin x}\,dx \\ &=\left[\frac{\sin 2nx}{n}\right]^{\frac \pi 2}_0 = 0\,\,\mathrm{(for\, all\,} n\neq0) \end{aligned}

Therefore, $I_n = I_{n-1}$ for all $n\neq 0$.

i.e. $I_1=I_2=I_3= \,\,...$

Now, evaluating $I_1$, \begin{aligned} I_1 &=\int_0^{\frac \pi 2}\frac {\sin(3x)}{\sin x}\,dx \\ &=\int_0^{\frac \pi 2}\frac{\sin 2x\cos x+\cos 2x\sin x}{\sin x}\,dx \\ &=\int_0^{\frac \pi 2}\frac{2 \sin x\cos^2 x+\cos^2x\sin x-\sin^3x}{\sin x} \,dx\\&=\int_0^{\frac \pi 2}3 \cos^2x-\sin^2x\,dx\\ &=\int_0^{\frac \pi 2}2 \cos^2x+\cos2x\,dx \\ &=\int_0^{\frac \pi 2}2 \cos 2x+1\,dx \\ &= \frac \pi 2 \end{aligned}

From above, we conclude that $I_n=\frac \pi 2$ for all non-negative $n$.

Therefore, $$\lim_{n\to\infty} 2I_n = \pi$$

• How did you assume that $n$ is an integer because without that you can't say that $\sin 2n\pi=0$ Commented May 8, 2018 at 14:25
• Huh... that’s a problem. My bad, I forgot about that. To be honest I’m not sure if the proof works without it. Commented May 8, 2018 at 14:31

With $$\frac{a^m-a^{-m}}{ a-a^{-1}}=a^{m-1} +a^{m-3} + \dots + a^{-m+1}$$ \begin{align}\int_0^\frac\pi2 \frac{\sin (2n+1)x}{\sin x} dx =&\int_0^\frac\pi2 \frac{e^{i(2n+1)x}-e^{-i(2n+1)x}}{e^{ix}-e^{-ix}} dx =\int_0^\frac\pi2 \sum_{k=0}^{2n} e^{2i(n-k)x}dx\\ \end{align} Note that $$Re\int_0^\frac\pi2 e^{2i (n-k) x}dx =0$$, except for $$k=n$$. Then \begin{align} \int_0^\frac\pi2 \frac{\sin (2n+1)x}{\sin x} dx = \int_0^\frac\pi2 1 \ dx=\frac\pi2 \end{align}

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 becomes simply becomes $0$.

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

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

By @PKBeam, since $$I_{n+1}-I_n=0$$ So $$I$$ does not depend on $$n$$, so we can write $$I_n=I_0=\int_{0}^{\pi/2} dx=\pi/2$$