Help evaluating a rational integral using residue

[TopCat]

Homework Statement

I need to compute \int_0^\infty \frac{dx}{x^3+a^3}.

Homework Equations

If f = g/h, then Res(f, a) = \frac{g(a)}{h'(a)}.

The Attempt at a Solution

In the first I've used a semicircular contour in the upper plane that is semi-circular around the pole at -a. So I calculate the integral to be

1/2 (2\pi iRes(f, -ae^i{2\pi/3}) - \pi i Res(f, -a)) = 1/2 (\frac{2\pi i}{3a^2 e^i{4\pi/3}}-\frac{\pi i}{3a^2}) = \cdot \frac{-\pi i}{3\sqrt{3}a^2+6a^2}. The 1/2 comes from those residues being the calculation of the integral over the whole real line, the first term comes from the integral over the entire contour, which contains the pole at -a\zeta_3, while the second term from the part of the contour around x=-a. The answer is listed as \frac{2\pi}{3\sqrt{3}a^2}, so I'm not sure where I went wrong.

 

[haruspex]

In the first I've used a semicircular contour in the upper plane that is semi-circular around the pole at -a.

Your integration path is not clear to me. Can you be more precise?
There seems to be an i missing in the locations of the complex poles.

 

[TopCat]

Your integration path is not clear to me. Can you be more precise?
There seems to be an i missing in the locations of the complex poles.

Sure. My contour is the upper half of a circle of radius r, the segment from -r to -a-e, the upper half of a circle of radius e around -a, and the segment from -a+e to r. I took the limits r -> infinity and e -> 0. The integral over the arc goes to zero, so I end up with the integral over the real line + pi*i*residue at -a = 2*pi*i*residue inside the contour.

I notice that I forgot to add the i into the exponent of zeta_3, I'll edit the original post. I did use it in my calculations.

 

[haruspex]

But the desired integral is only over the positive real line.

 

[TopCat]

But the desired integral is only over the positive real line.

That's why I multiplied by 1/2. But now that I think about it, that wouldn't work since the function isn't symmetric about the y-axis, right?

 

[haruspex]

Right.

 

[TopCat]

I don't see which contour would be best to use in this situation.

 

[jackmell]

I don't see which contour would be best to use in this situation.

Tell you what, first get it working for say a=2 by introducing a branch cut along the real axis and considering:

\oint \frac{\log(z)}{z^3+8}dz

around a key-hole contour with the key-hole along the positive real axis. One minor issue is the residue in the fourth quadrant. Note the analytic-extension of \log(z) over that contour into that quadrant is actually \text{Log}(z)+2\pi i right? Ok, get that one then generalize it to any a.

Edit:

Afraid I made this more complicated than it needed to be: Easier probably to integrate directly 1/(z^3+8) over a 2\pi/3 wedge contour bordering the positive real axis and in that way, only encircle one pole in the first quadrant and not even need to worry about dealing with the log branch-cut. Sorry.

 

[haruspex]

integrate directly 1/(z^3+8) over a 2\pi/3 wedge contour bordering the positive real axis and in that way, only encircle one pole in the first quadrant

... and the trick then is to express the integral over the line r e^2πi/3 as a constant times that over (0, ∞) portion, yes?

 

[jackmell]

... and the trick then is to express the integral over the line r e^2πi/3 as a constant times that over (0, ∞) portion, yes?

Yes:

\mathop\int\limits_{C} \frac{1}{z^3+8}dz=\int_{\infty}^0 \frac{1}{r^3 e^{6\pi i/3}+8} e^{2\pi i/3} dr

You can see how to get that right? Hope that's not an insulting question to ask.

Then surely, I think, we can do smaller wedges to figure out:

<br /> \int_0^{\infty} \frac{1}{z^n+a^n}dz

You may wish to look at that one to get the general principle down.