# $\lim_{n\to\infty}\int_{-\infty}^{\infty}{\sin(n+0.5)x\over \sin(x/2)}\cdot{\mathrm dx\over 1+x^2}=\pi\cdot{e+1\over e-1}$

How can we show that

$$\lim_{n\to\infty}\int_{-\infty}^{\infty}{\sin(n+0.5)x\over \sin(x/2)}\cdot{\mathrm dx\over 1+x^2}=\pi\cdot{e+1\over e-1}\tag1$$

$(1)$, substitution doesn't work, either integration by parts.

We know $(2)$

$$\int_{-\infty}^{\infty}{\mathrm dx\over 1+x^2}=\pi\tag2$$

$${\sin(n+0.5)x\over \sin(x/2)}={\sin(nx)\cos(x/2)+\sin(x/2)\cos(nx)\over \sin(x/2)}\tag3$$

Simplified to

$$=\sin(nx)\coth(x/2)+\cos(nx)\tag4$$

$$\lim_{n\to\infty}\int_{-\infty}^{\infty}\sin(nx)\cot(x/2)\cdot{\mathrm dx\over 1+x^2}+\int_{-\infty}^{\infty}\cos(nx)\cdot{\mathrm dx\over 1+x^2}=\pi\cdot{e+1\over e-1}\tag5$$

$$\lim_{n\to\infty}\int_{-\infty}^{\infty}\sin(nx)\cot(x/2)\cdot{\mathrm dx\over 1+x^2}+{\pi\over e^n}=\pi\cdot{e+1\over e-1}\tag6$$

$$\lim_{n\to\infty}\int_{-\infty}^{\infty}\sin(nx)\cot(x/2)\cdot{\mathrm dx\over 1+x^2}=\pi\cdot{e+1\over e-1}\tag7$$

I am not sure how to continue

• do you know the method of stationary phase? – tired Feb 3 '17 at 20:20

Here is one quick way. Note that

$${\sin(n+0.5)x\over \sin(x/2)}$$

is a Dirichlet kernel and we may use the identity (which can easily be proven)

$${\sin(n+0.5)x\over \sin(x/2)} = 1 + 2\sum_{k=1}^n \cos (kx)$$ to rewrite the integral as

\begin{align} I&= \lim_{n\to\infty}\int_{-\infty}^{\infty}{\sin(n+0.5)x\over \sin(x/2)}\cdot{\mathrm dx\over 1+x^2}=\lim_{n\to\infty}\int_{-\infty}^{\infty}\left[1 + 2\sum_{k=1}^n \cos (kx)\right]\cdot{\mathrm dx\over 1+x^2}\\ &=\int_{-\infty}^{\infty}{\mathrm dx\over 1+x^2} + 2\sum_{k=1}^{\infty }\int_{-\infty}^{\infty}{\cos (kx) \over (1+x^2)} \end{align}

The latter integral can be evaluated using residues to get

$$I = \pi + 2\pi \sum_{k=1}^{\infty }e^{-k}=\pi + \frac{2\pi}{e-1}=\pi \frac{e+1}{e-1}$$