Efficiently Solving Integrals with Residues | Integral Calculus Homework

[nicksauce]

Homework Statement

Evaluate $$\int_0^{\infty} \frac{dx}{1+x^{100}}$$

Homework Equations

$$\int^{\infty}_{-\infty} \frac{P(x)dx}{Q(x)} = 2\pi i\sum_{\textnormal{res}}\frac{P(z)}{Q(z)}$$ in the Upper half plane.

The Attempt at a Solution

I really can't be expected to calculate the residue of this function some 50 times can I? There must be some trick I am missing. Any hints?

 

[Hurkyl]

At the very least, have you written down what the summation would be?

 

[nicksauce]

Ok so I need to sum all the residues of
$$\frac{1}{1+z^{100}}$$ in the upper half plane. There are simple poles at $$z=e^{i\pi/100},e^{3i\pi/100},e^{5i\pi/100}...e^{99i\pi/100}$$.

The residue at the first pole is
$$\lim_{z\rightarrow e^{i\pi/100}} \frac{z-e^{i\pi/100}}{1+z^{100}}$$
$$= \frac{1}{100e^{99i\pi/100}}$$

The residue at the nth pole (the first being the zeroth) will be:
$$\frac{1}{100e^{(1+2n)i\pi*99/100}}$$

Does that look okay so far?

 

[Hurkyl]

Does that look okay so far?

I expected something like that; why'd you stop there?

 

[nicksauce]

Okay so then I can get

Sum of residues =
$$\frac{1}{100}\sum_{n=0}^{n=49}e^{-(1+2n)i\pi99/100}$$

Any way to do this analytically?

 

[Hurkyl]

Okay so then I can get

Sum of residues =
$$\frac{1}{100}\sum_{n=0}^{n=49}e^{-(1+2n)i\pi99/100}$$

Any way to do this analytically?

Yes; this is a kind of sequence you're quite familiar with. How are consecutive terms related?