Integration by Residue Calculus

[stephenkeiths]

Homework Statement

I'm trying to evaluate the integral

I(a)=\int\frac{cos(ax)}{x^{4}+1}
from 0 to ∞

Homework Equations

To do this I'm going to consider the complex integral:

J=\oint\frac{e^{iaz}}{z^{4}+1}

Over a semi-circle of radius R in the upper half plane, then let R-->∞

There are 2 curves, the straight curve along the real axis, and the semi-circular arc.

The Attempt at a Solution

Along the first curve (along the real axis) the line integral is just

I_{1}=\int\frac{e^{iax}}{x^{4}+1}

from 0 to infinity

Now the real part of this is 2*I(a)

The second line integral goes to zero. So 2*I(a)=Re(J).

So I need to evaluate J by residue calculus and take the real part to get I(a).

So there are 2 simple poles inside of the semi-circle, namely
z=e^{i\frac{\pi}{4}} and z=e^{i\frac{3\pi}{4}}

So J=2i\pi(Res(f,e^{i\frac{\pi}{4}})+Res(f,e^{i\frac{3\pi}{4}}))

But I'm having trouble computing the Residues.

I know that Res(f,e^{i\frac{\pi}{4}})=\frac{e^{iaz}}{4z^{3}} z=e^{i\frac{\pi}{4}}

But I'm struggling with the exponential to an exponential, I can't get a "clean" answer. I'm not sure if there's a better way I should go about this. Or if there's something I'm doing wrong.

Please Help!

 

[Mute]

But I'm having trouble computing the Residues.

I know that Res(f,e^{i\frac{\pi}{4}})=\frac{e^{iaz}}{4z^{3}} z=e^{i\frac{\pi}{4}}

Are you sure about that result? I don't think there should be a z in the numerator. It looks like you calculated the residue using L'Hopital's rule, is that right? i.e.,

$$\lim_{z\rightarrow z_1} (z-z_1)\frac{e^{i\alpha z}}{z^4+1} = \lim_{z\rightarrow z_1} \frac{(z-z_1)'}{(z^4+1)'}e^{i\alpha z}.$$
When you apply the rule, you will get $4z_1^3$ in the denominator, but your numerator should just be a constant.

But I'm struggling with the exponential to an exponential, I can't get a "clean" answer. I'm not sure if there's a better way I should go about this. Or if there's something I'm doing wrong.

Please Help!

You can always write the roots in cartesian form, $z = a + ib$.

 

[stephenkeiths]

Thanks for replying

Thats what I got for my residue, Sorry that z is just saying plug in e^i...

I try writing it in that form (a+bi), but its so messy!

What I get is

J=\frac{\pi\sqrt{2}}{2}e^{-a\frac{\sqrt{2}}{2}}(cos(a\frac{\sqrt{2}}{2})+sin(a\frac{\sqrt{2}}{2}))

Does anyone else think I'm crazy? I really feel like somethings wrong here!

 

[Dick]

Thanks for replying

Thats what I got for my residue, Sorry that z is just saying plug in e^i...

I try writing it in that form (a+bi), but its so messy!

What I get is

J=\frac{\pi\sqrt{2}}{2}e^{-a\frac{\sqrt{2}}{2}}(cos(a\frac{\sqrt{2}}{2})+sin(a\frac{\sqrt{2}}{2}))

Does anyone else think I'm crazy? I really feel like somethings wrong here!

If that's your answer for the whole integral, I'll agree with it, at least for the case of a>=0. If you aren't give that, I'd write |a| instead of a. I'll agree it's not a "really clean" answer. But it's about as clean as you are going to get.

 

[stephenkeiths]

Ya I only need to solve for a>0

Thanks! It gives me some confidence!