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=limt→∞∫t01t(t−t33!+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=π2−In=∫π/20f(x)sin(2n+1)x dx
where f(x)=1sinx−1x.
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+1∫π/20f′(x)cos(2n+1)x dx=O(1/n).
Hence In→π/2 and we conclude that ∫∞0sinxxdx=limn→∞In=π2.
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