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Integrating power of a cosine times a complex exponential

  1. Jan 18, 2014 #1

    bcf

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    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

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    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
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