Evaluate the integral using the residue theorem and its applications.

[StumpedPupil]

Homework Statement

the integral from negative infinity to positive infinity: z^4/(1 + z^8)dz

Homework Equations

The residue theorem: <http://en.wikipedia.org/wiki/Residue_theorem>.

The Attempt at a Solution

I found the 8th roots of z^8 = -1, which are e^(πiz), where z =1/8, 3/8, 5/8, 7/8, 9/8 11/8, 13/8 15/8.

So the denominator is (z - z0)(z - z1)(z - z2) . . . (z - z7) and we can calculate the residue of each pole. Each pole has order 1, so the residue at each pole is just cauchy's formula at the pole. Then the integral is just the sum of the 8 residues multiplied by 2πi.

This seems more complicated then it should be. Can anyone tell me if I am doing this right because something seems off. Thank you.

 

[Dick]

It is complicated but it should work. But you don't want to add ALL of the residues. You'll get zero. You want to draw a CLOSED contour that includes the real axis. That contour is only going to include four of the poles, isn't it?

 

[HallsofIvy]

Homework Statement

the integral from negative infinity to positive infinity: z^4/(1 + z^8)dz

Homework Equations

The residue theorem: <http://en.wikipedia.org/wiki/Residue_theorem>.

The Attempt at a Solution

I found the 8th roots of z^8 = -1, which are e^(πiz), where z =1/8, 3/8, 5/8, 7/8, 9/8 11/8, 13/8 15/8.

You are using "z" with two different meanings here! Please don't do that!

So the denominator is (z - z0)(z - z1)(z - z2) . . . (z - z7) and we can calculate the residue of each pole. Each pole has order 1, so the residue at each pole is just cauchy's formula at the pole. Then the integral is just the sum of the 8 residues multiplied by 2πi.

This seems more complicated then it should be. Can anyone tell me if I am doing this right because something seems off. Thank you.

Your integral is along the real line from negative infinity to infinity. What contour are you going to use to get that? I would recommend integrating from -R to R on the real line then along the semicircle |z|= R with imaginary part positive. As soon as R> 1 four of those roots will be inside the contour and the integral around the entire contour will be the sum of those four residues. Finally let R go to infinity. You will have to determine what happens to the integral on the semi-circle in that case.

 

[StumpedPupil]

Yes, only four poles are in the upper half plane, namely the values for the variable w, which I mistakenly labeled as a z when z already existed. So I will integrate over the semi circle for very large R in the upper half plane for the poles z = e^(πiw), where w =1/8, 3/8, 5/8 and 7/8. Thank you for all your help.