I'm trying to show that
∫+∞0sin(x)x(x2+1)dx=π2(1−1e)
using Jordan's lemma and contour integration.
MY ATTEMPT: The function in the integrand is even, so I have:
∫+∞0sin(x)x(x2+1)dx=12∫+∞−∞sin(x)x(x2+1)dx
There is a simple pole at z=0 and poles at z=+i,z=−i.
A method in the chapter I am working on (Ablowitz & Fokas sections 4.2 & 4.3) usually considers the integral
∫+∞−∞eixx(x2+1)dx=2πiRes(eixx(x2+1),z=i,−i,0)
But I am not sure if this will work, instead another example builds a contour Cr+Ce+(−R,−e)+(R,e) which avoids the poles and thus integrating over that yields zero and helps me get my answer. Unfortunately, this attempt does not give me the right value either.
Do any of you integration whizzes out there have anything for me? Many thanks.
Answer
Worked this out, it's a combination of both methods I mention. You need to go around the contour Cr+Ce+(−R,−e)+(R,e). By the only theorem on section 4.3 of Ablowitz & Fokas the integral around Ce will go to iπ. By the approach I mentioned in the first part of my attempt, the integral around CR will go to 2πi(−12e) adding these two solutions up, and remembering to divide by 2 (notice the integrand is even) yields the answer.
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