Thursday, June 13, 2019

calculus - Evaluating the integral inti0nftyfracsinxx,mathrmdx=fracpi2?


A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: 0sinxxdx=π2


Well, can anyone prove this without using Residue theory. I actually thought of doing this: 0sinxxdx=lim but I don't see how \pi comes here, since we need the answer to be equal to \dfrac{\pi}{2}.



Answer



Here's another way of finishing off Derek's argument. He proves \int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2. Let I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx= \int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx. Let D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx where f(x)=\frac1{\sin x}-\frac1x. We need the fact that if we define f(0)=0 then f has a continuous derivative on the interval [0,\pi/2]. Integration by parts yields D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n). Hence I_n\to\pi/2 and we conclude that \int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.


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