How to use the residue theorem when I have an exponential as the pole?

[laura_a]

Homework Statement

Q. Use residues and the contour shown (where R > 1) to establish the integration formula

$$\int^{\infty}_{0} \frac{dx}{x^3+1} = \frac{2\pi}{3 \sqrt{3}}$$

The given contour is a segment of an arc which goes from R (on the x-axis) to Rexp(i*2*pi/3)

Homework Equations

The Attempt at a Solution

For the function [itex] f(x) = \frac{dx}{x^3+1} [/tex]
I have worked out that there is only only one pole inside the contour (R >1) which is $$x=e^{\frac{i\pi}{3}}$$

So the residue would be

2 * pi * i * Res_($$x=e^{\frac{i\pi}{3}}$$) f(x)

But I'm not sure how to do that residue as I've never calculated one with an e in it before... any suggestions? :) Thanks

Homework Statement

p.s sorry I couldn't get the LaTex to show up, someone told me how to do it, but that is obviously not it...

 

[BSMSMSTMSPHD]

The opening tags shouldn't have a slashes.

 

[Dick]

To deal with the part of the contour along the complex line substitute t*exp(i*2*pi/3) into the integral. You should be able to convince yourself that this part of the contour has been cleverly picked to have a simple relation to the integral along the x axis. If we call the integral along the x-axis I, then the integral along this line going outward is I*exp(i*2*pi/3). Now convince yourself you can ignore the contribution from the arc in the limit as R->infinity. So the total integral along the contour is I-I*exp(i*2*pi/3). (The minus sign coming because we are integrating counterclockwise around the region). Now equate this to what you get from the residue theorem (it's a simple pole, just like in the 1/(x^2+1) case) and solve for I.

 

[Hurkyl]

But I'm not sure how to do that residue as I've never calculated one with an e in it before... any suggestions? :)

Er, it's just a number, like any other. What's giving you problems?


Incidentally, you have the wrong tags for LaTeX. You want
[tex]YourLatexHere[/tex]

 

[Dick]

The e doesn't make any difference. It's still lim(z->z0) (z-z0)*f(z). Factor the cubic as (z-z0)*(z-z1)*(z-z2) where the zi's are the three cube roots of -1.