1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrating power of a cosine times a complex exponential

  1. Jan 18, 2014 #1


    User Avatar

    1. The problem statement, all variables and given/known data
    Consider the inner product
    $$\frac{1}{2\pi}\int_0^{2\pi} \left(\frac{3}{5 - 4\cos(x)}\right) e^{-ikx} dx, \quad k \in \mathbb{Z}, \quad x \in \mathbb{R}.$$

    2. Relevant equations
    Is there a method to solve this without using the residue theorem, e.g. integration by parts, a clever substitution, etc.?

    3. The attempt at a solution
    One attempt: First rewrite the rational function using geometric series:
    $$\frac{3}{5 - 4\cos(x)} = \frac{3}{5}\sum_{n = 0}^\infty (\frac{4}{5}\cos(x))^n$$.
    This gives the integral (assuming we can swap the series and integration?)
    $$\frac{1}{2\pi}\frac{3}{5} \sum_{n = 0}^\infty (\frac{4}{5})^n\int_0^{2\pi} \cos^n(x) e^{-ikx} dx.$$
    Now, rewrite the cosine function in terms of complex exponentials, then use binomial theroem:
    $$ \cos^n(x) = (\frac{1}{2}(e^{ix} + e^{-ix}))^n = (\frac{1}{2})^n \sum_{j = 0}^n {n \choose j} e^{ijx}e^{-i(n-j)x}.$$
    Sub this into the integral and apply linearity, so we're now have the expression
    $$\frac{1}{2\pi}\frac{3}{5} \sum_{n = 0}^\infty (\frac{4}{5})^n (\frac{1}{2})^n \sum_{j = 0}^n {n \choose j} \int_0^{2\pi} \exp(i(2j - n - k)x) dx.$$
    At least now I have an integral I know how to evaluate. The integral is zero unless 2j - n - k = 0, in which case it is simply 2*pi. The question is how to write the final expression, knowing that all terms are zero except when 2j -n = k?
  2. jcsd
  3. Jan 18, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    You could have started out rewriting the cosine as complex exponentials at the start - or writing the exponential as a sum of trig functions and using a table of trig identities maybe?

    May work to represent the integral as a differential equation.

    Like you, I'd be tempted to try some sort of series expansion - that's mostly useful where I am allowed to take an approximation though.

    Per your question: looks like the Kronecker delta is your friend there.
    would be: $$\frac{1}{2\pi}\int_0^{2\pi} \exp \big[i (2j-n-k)x\big]\;\text{d}x = \delta_{2j,n+k}$$... or something?
    Last edited: Jan 18, 2014
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted