Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that the function $\csc(x/2)-2/x$ is integrable on $(0,\pi)$. In fact, prove that it is bounded. In fact, prove that it tends to zero as $x\to0$. Use this to show that $$\lim_{N\to\infty}\int_0^\pi\left(\frac1{\sin\frac{x}2}-\frac2x\right)\sin\left((N+\frac12)x\right)dx=0$$ Then prove that $$\lim_{N\to\infty}\int_0^\pi\frac{\sin(N+\frac12)x}xdx=\pi/2$$ Finally, prove that $$\int_0^\infty\frac{\sin x}xdx=\pi/2$$

One observation is $\sum_{n=-N}^Ne^{inx}=\frac{\sin(N+\frac12)x}{\sin\frac{x}2}$, and the integral of this from $0$ to $2\pi$ is $2\pi$, since the integral $\int_{0}^{2\pi}e^{inx}dx=0$ if $n\neq0$.

share|cite|improve this question
Funny timing. I posted this argument a few days ago. The full details are in my blog, here. – Andrés E. Caicedo Dec 4 '13 at 6:39

First note that as $x \to 0^{+}$ the function $$f(x) = \dfrac{1}{\sin\left(\dfrac{x}{2}\right)} - \frac{2}{x}$$ tends to a definite limit $0$ and hence can be assumed continuous in $[0, \pi]$. Therefore $f(x)$ is Riemann-integrable on interval $[0, \pi]$. It now follows by Riemann-Lebesgue Lemma (related to coefficients of Fourier series of $f(x)$, a proof is available in Tom M. Apostol's "Mathematical Analysis", 2nd Ed., Page 313) that $$\lim_{N \to \infty}\int_{0}^{\pi}f(x)\sin(Nx + b)\,dx = 0$$ This settles the hard part of the question. The limit of $f(x)$ as $x \to 0^{+}$ is calculated as follows:

$\displaystyle \begin{aligned}\lim_{x \to 0^{+}}f(x) &= \lim_{x \to 0^{+}}\dfrac{1}{\sin\left(\dfrac{x}{2}\right)} - \frac{2}{x}\\ &= \lim_{x \to 0^{+}}\dfrac{x - 2\sin(x/2)}{x\sin(x/2)}\\ &= \lim_{x \to 0^{+}}\dfrac{x - 2\sin(x/2)}{x\dfrac{\sin(x/2)}{x/2}\cdot(x/2)}\\ &= 2\lim_{x \to 0^{+}}\dfrac{x - 2\sin(x/2)}{x^{2}}\\ &= 2\lim_{x \to 0^{+}}\dfrac{1 - \cos(x/2)}{2x}\text{ (by L'Hospital's Rule)}\\ &= \lim_{x \to 0^{+}}\dfrac{2\sin^{2}(x/4)}{x}\\ &= 2\lim_{x \to 0^{+}}\dfrac{\sin^{2}(x/4)}{(x/4)^{2}}\cdot\frac{(x/4)^{2}}{x} = 0\end{aligned}$

The observation which you have made about representing $\sin(N + 1/2)x$ as a sum can help you out in showing that $$\int_{0}^{\pi}\dfrac{\sin\left(N + \dfrac{1}{2}\right)x}{\sin(x/2)} = \pi$$ Using this result together with the earlier established limit $$\lim_{N \to \infty}\int_{0}^{\pi}\left(\dfrac{1}{\sin(x/2)} - \frac{2}{x}\right)\sin\left(N + \frac{1}{2}\right)x\,dx = 0$$ gives us $$\lim_{N \to \infty}\int_{0}^{\pi}\dfrac{\sin\left(N + \dfrac{1}{2}\right)x}{x}\,dx = \frac{\pi}{2}$$The last part of the question can be easily deduced by putting $(N + 1/2)x = t$ in the above integral.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.