Real integral using complex variables

[Grothard]

Homework Statement

\int_{0}^{\infty }\frac{dx}{1+x^{5}}


The attempt at a solution

This is for my complex variables class, so I have been trying to compute it using residues. I noticed that if we extend it to the complex plane and integrate over the edges of a half disk or a quarter disk (or even 3/4 of a disk) of radius R, as R approaches infinity the circular curve part approaches zero. This means that

\int_{0}^{\infty }\frac{dx}{1+x^{5}} + \int_{0}^{\infty }\frac{dx}{1+ix^{5}} = Res(enclosed)

and

\int_{0}^{\infty }\frac{dx}{1+x^{5}} + \int_{0}^{\infty }\frac{dx}{1-ix^{5}} = Res(enclosed)

The residues aren't hard to compute, but I don't know what to do with the other integral

 

[Dick]

Well, if you add those two equations you can put the complex integrals together to get the integral of 2/(1+x^10). And that you can do with the residue theorem. There may be an easier way to do this, but it's not coming to me right now.

 

[lawsofform]

Or, you can add those two equations, then use the fact that 1 + (-x)^5 = 1 - x^5 to make the integral over (-inf,inf).

 

[Gib Z]

Instead of a semi-circle, or a quarter, try integrating along the sector with angle \frac{2\pi}{5}. Then the integral along the other straight side should look familiar to you, and the contribution from the arc still goes to zero in the limit.

EDIT: In fact, it is instructive to prove something more general: \int^{\infty}_0 \frac{1}{1+x^n} dx = \frac{ \pi \csc (\pi/n)}{n}, where you use the sector with angle 2\pi/n.

 

[Grothard]

That should work, thanks!

 

[Grothard]

Actually, I am still running into some problems.

Along the other straight side the x term can be expressed as x = re^{2\pi/n} where r is the distance from the origin. Clearly \frac{1}{1+x^{n}} = \frac{1}{1+r^{n}}, so integrating the side from 0 to R with respect to r will result in the same integral as the other side.
The problem is that since this is an oriented closed curve, we should be integrating the other way (from R to 0), which would make the integral negative and the two sides would cancel.
Aside from that, something is wrong with my calculation of the residue at e^{i\frac{\pi }{5}}. I factor the polynomial and remove the \frac{1}{(x - e^{i\frac{\pi }{5}})}, but I end up with this:
http://www.wolframalpha.com/input/?...i/5)-e^(i*7*pi/5))*(e^(i*pi/5)-e^(i*9*pi/5))]
which I can't see becoming the form of the answer you mentioned

 

[Gib Z]

The integrals don't quite cancel (that would be useless for us!). So the integral along the real axis is \int^R_0 \frac{1}{1+x^n} dx and no other work needed, its already in a kind of "simple" form we want. The integral along the other side (account for the negative factor in changing directions) is - \int_{\gamma} \frac{1}{1+z^n} dz where the path is \gamma(t) = t \cdot e^{2i\pi/n} and t varies from 0 to R.

You need to apply this formula correctly:
If the contour C(t) has t range from a to b, then \int_{C} f(z) dz = \int^b_a f(C(t)) C'(t) dt. It won't be precisely the same integral as along the real axis, but it will be some multiple of it (because of the C'(t) term), which is good.

As for your residue calculation (I'll keep on being general using n instead of n=5 ), for some reason I see people going through a heap of grief over computing simple residues. Many things you learned in calculus are still valid here, such as l'hospital's rule! It makes most residue calculations a piece of cake. For example here, the residue at z= e^{i\pi/n} is given by this limit: L = \lim_{z\to e^{i\pi /n}} \frac{ z - e^{i\pi/n} }{1+z^n}. This is clearly of the 0/0 form (if the numerator and the denominator didn't tend to 0, this wouldn't be a residue calculation!) so just using l'hospital's rule simplifies it greatly: L = \lim_{z\to e^{i\pi/n} } \frac{1}{nz^{n-1}} = \frac{1}{n} \lim_{z\to e^{i\pi/n} } \frac{z}{z^n}. I wrote the last inequality in that way because z^n just tends to -1.