Problem 8

Exercise 11.8 (Series for a Complex Integral)   Prove that

$$\frac{1}{2\pi}\int_0^{2\pi} e^{\cos\theta}d\theta = \sum_{n=0}^\infty \frac{1}{(n!2^n)^2}.$$ (11.42)

Proof. Substitute $z=e^{i\theta}$, so that

$$\cos\theta = \frac{1}{2}(z + 1/z) \quad d\theta = \frac{dz}{iz}$$ (11.43)

Then the above integral can be interpreted as a winding number about the unit circle.

$$\frac{1}{2\pi}\int_0^{2\pi} e^{\cos\theta}d\theta= \frac{1}{2\pi}... .../2(z+1/z)}\frac{dz}{iz}= \frac{1}{2\pi i}\int_\gamma \frac{e^{1/2(z+1/z)}}{z}dz$$ (11.44)

To summon the appropriate poles to integrate over, expand

$$e^{1/2z} = \sum_{k=0}^\infty \frac{1}{k!2^kz^k}$$ (11.45)

Then

$$\frac{e^{1/2(z+1/z)}}{z} = \sum_{k=0}^\infty \frac{1}{k!2^kz^k} \... ...um_{k=0}^\infty \frac{1}{k!2^kz^{k+1}} \left(1+z/2 + (z/2)^2/2! + \cdots\right)$$ (11.46)

Integrating ignores all the terms which are not a multiple of $1/z$, so we note that the coefficient of $1/z$ equals

$$\frac{(1/2)^k/k!}{k!2^k}=\frac{1}{(k!2^k)^2}$$ (11.47)

by finding the numerator from the expansion of $e^{z/2}$ and the denominator from the denominator in the expansion involving $e^{1/2z}$. Therefore

$$\frac{1}{2\pi}\int_0^{2\pi} e^{\cos\theta}d\theta =\sum_{k=0}^\infty \frac{1}{(k!2^k)^2}.$$ (11.48)

$\qedsymbol$

For a version of this problem with an arbitrary scaling factor, see Problem 5.10.6 of [13].