Prove that $ \int _{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$

Question

Prove $$\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$$ Start with the fact that the integral of $D_{N}(\theta)$ equals $2\pi$, and note that the difference $$\frac{1}{\sin(\theta/2)}-\frac{2}{\theta}$$ is continuous on $[-\pi,\pi]$. Apply the Riemann-Lebesgue lemma. $$D_{N}(x)=\sum_{-N}^{N}e^{inx}=\frac{\sin((N+1/2)x)}{\sin(x/2)}$$ Now I can only think to apply the Riemann-Lebesgue lemma to the function $$\frac{1}{\sin(\theta/2)}-\frac{2}{\theta}$$ To get : $$\lim_{n\to \infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}(\frac{e^{-in\theta}}{\sin(\theta/2)}-\frac{2e^{-in\theta}}{\theta})d\theta\to0$$But I don't know the steps to prove this.

Answer 1

Note that $$\int_{-\pi}^\pi D_N(t)\,dt=2\pi.$$ Use that $$\int_{-\pi}^\pi\sin((N+1/2)t)\left(\frac1{\sin t/2}-\frac 2t\right)\,dt\to0$$ as $N\to\infty$ (Riemann-Lebesgue) to deduce that $$\lim_{N\to\infty}\int_{-\pi}^\pi\frac{\sin((N+1/2)t)}{t}\,dx=\pi.$$

Answer 2

Consider the Laplace transform $$ F(s)=\int_0^{\infty}e^{-sx}\frac{\sin x}{x}dx. $$ Your integral is $\lim_{s\downarrow 0}F(s)$. For $s > 0$, $$ F'(s)=\int_{0}^{\infty}e^{-sx}\sin xdx \\ = \Im \int_0^{\infty}e^{-sx}e^{ix}dx \\ = \Im \int_0^{\infty}e^{-x(s+i)}dx \\ = \Im \frac{1}{s+i}=\Im\frac{s-i}{1+s^2} \\=-\frac{1}{1+s^2} $$ Therefore $F(s)=-\tan^{-1}(s)+C$ for some constant $C$. Because $F(\infty)=0$, then $C=\frac{\pi}{2}$. So $$ \int_0^{\infty}e^{-sx}\frac{\sin x}{x}dx = \frac{\pi}{2}-\tan^{-1}(s). $$ Letting $s\downarrow 0$ gives $$ \int_0^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}. $$