Efficiently Solving Integrals with Residues | Integral Calculus Homework

  • Thread starter Thread starter nicksauce
  • Start date Start date
  • Tags Tags
    Integral
nicksauce
Science Advisor
Homework Helper
Messages
1,270
Reaction score
7

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?
 
Physics news on Phys.org
At the very least, have you written down what the summation would be?
 
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?
 
nicksauce said:
Does that look okay so far?
I expected something like that; why'd you stop there?
 
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?
 
nicksauce said:
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?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top