Solving the integral ∫∞0sinxx dx=π2?

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

∫∞0sinxx dx=π2

Well, can anyone prove this without using Residue theory. I actually thought of doing this:

∫∞0sinxx dx=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.