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Question:

Show that $\int_0^{2\pi} e^{\cos\theta} \ d\theta=\sum_{n=0}^{\infty} \frac{1}{(n!2^n)^2}$

My attempt:

We are learning about laurent series and their integral representation. I think I have to make the substitution $z=e^{i\theta}$ and convert the integral into $$\frac{1}{2\pi i}\int_C\frac{e^{(z+\frac{1}{z})/2}}{z} \ dz,$$ where $C$ is the unit circle in the complex plane. I don't know what to do from here though.

user210552
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    It's not obvious the complex analysis route would be the fastest. Any reason not to write out a Taylor series in $\cos\theta$ and then use formulae for the integrals $\int_0^{2\pi} \cos^n\theta\ d\theta\ $? – Simon S Oct 19 '15 at 21:36
  • hmmm...the reason why I didn't try that is because this is a complex analysis class and my professor insists that complex variable techniques will yield elegant solutions on all the problems – user210552 Oct 19 '15 at 21:39
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    What to do from where you have come: Multiply the Laurent series $$\Biggl( \sum_{n = 0}^\infty \frac{z^n}{2^n\cdot n!}\Biggr)\cdot \Biggl(\sum_{k = 0}^\infty \frac{z^{-k}}{2^k\cdot k!}\Biggr),$$ collect the terms with exponent $0$ in the product (since all other terms will, when multiplied with $z^{-1}$ and integrated over the circle, produce $0$). If you don't have $\int_0^{2\pi} \cos^n\theta,d\theta$ memorised, that is quicker than the real-methods way. If you have $\int_0^{2\pi} \cos^n \theta,d\theta$ memorised, the real method is probably a little quicker. – Daniel Fischer Oct 19 '15 at 21:56
  • Now that I see Daniel's solution I like the complex approach – Simon S Oct 19 '15 at 21:58
  • I think you forgot a factor of $2\pi$ at the front.

    See http://www.wolframalpha.com/input/?i=integrate+exp%28cos%28x%29%29+from+0+to+2pi

     – Almentoe Oct 19 '15 at 22:28
  • A related question. – Lucian Oct 20 '15 at 03:56

1 Answers1

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$$\int_0^{2\pi} \exp(\cos\theta) \ d\theta = $$

$$ \int_{|z|=1}\frac{-i\exp(z+\frac{1}{z})}{z}dz = 2 \pi i \text{Res}(f,0) $$

Because $0$ is the only singularity of $f(z) := -iz^{-1}\exp(z+\frac{1}{z}) $

$\text{Res}(f,0) = $ coefficient of $z^{-1}$ in the Laurent expansion of $f$, let's find this expansion:

$f(z) = \sum_{n=0}^{\infty}\frac{-i(z+z^{-1})^n}{2^nz} = \sum_{n=0}^{\infty} \sum_{j=0}^{n} -i\frac{n!}{j!(n-j)!2^{2n}} z^{n-2j-1} $

So the $z^{-1}$ is determined by the equation $n-2j-1 = -1$, so $n = 2j$, let's plug this in:

$$ \int_0^{2\pi} \exp(\cos\theta) \ d\theta = 2\pi \sum_{n=0}^{\infty} \frac{(2j)!}{j!j!(2j)!2^{2j}} = 2\pi \sum_{n=0}^{\infty} \frac{1}{(j!)^22^{2j}} $$ as desired.

I have left the algebra for you to do. I think you left out a factor of $2\pi$ in the answer, see http://www.wolframalpha.com/input/?i=integrate+exp%28cos%28x%29%29+from+0+to+2pi and http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html

Almentoe
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