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=limtt01t(tt33!+t55!+)dt

but I don't see how π comes here, since we need the answer to be equal to π2.



Answer



Here's another way of finishing off Derek's argument. He proves π/20sin(2n+1)xsinxdx=π2.

Let In=π/20sin(2n+1)xxdx=(2n+1)π/20sinxxdx.
Let Dn=π2In=π/20f(x)sin(2n+1)x dx
where f(x)=1sinx1x.
We need the fact that if we define f(0)=0 then f has a continuous derivative on the interval [0,π/2]. Integration by parts yields Dn=12n+1π/20f(x)cos(2n+1)x dx=O(1/n).
Hence Inπ/2 and we conclude that 0sinxxdx=limnIn=π2.


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