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=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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