Contour integration problem-(sinx/x)^2

[outhsakotad]

contour integration problem--(sinx/x)^2

Homework Statement

I am to evaluate the integral of (sinx/x)^2 from -infinity to +infinity.

Homework Equations

The Attempt at a Solution

I drew my contour as a large half circle in UHP, and this contour then includes the singularity at the origin.

Using a trig identity, I can rewrite sin^x : 0.5(1-cos(2x))

Since we're working in the UHP, I then rewrite the integral: Real part of 0.5*int((1-e^(2iz))/z^2)dz from -inf to +inf

There is a second order pole at the origin. The residue here is a-= 0.5*(d/dz){(z^2*(1-e^(2iz)))/z^2)} evaluated as z-->0, = 0.5*-2i*e^(2i*0)=-i.

Since the part enclosing the singularity is a half circle, the integral should be pi*i*a-=pi. But the answer is supposed to be pi/2.

I apologize for the lack of LaTeX. Could somebody please give me a hint as to what I'm doing wrong? Thanks.

 

[outhsakotad]

I now see that the error could possibly be that in converting the cosine to exponential form, I missed a factor of 1/2. But how come I have solved other problems correctly without that factor? See for example, the problem right before "Solution 13.18" on this document: http://www.cacr.caltech.edu/~sean/applied_math.pdf

 

[jackmell]

That's just a simple pole. Expand it as a power series to see that.

 

[outhsakotad]

Ah, okay, indeed it is a simple pole. But when I do the problem again, I still get pi. See the attached pdf.

 

[outhsakotad]

Here's the pdf:

 

[jackmell]

I get pi too and that's what Mathematica gives also.

 

[outhsakotad]

Hmmm. Arfken & Weber problem 7.1.12 asks me to prove that it's pi/2. Possible typo?

 

[vela]

In my copy of Arfken (3rd edition), it's problem 7.2.12, and the limits on the integral are from 0 to infinity.

 

[outhsakotad]

It must be a typo then. In the 6th edition, the limits are from -infinity to +infinity.