Combinatorics, Contour Integrals, and Bessel Functions … oh my?!

Here’s an interesting exercise that can be solved using complex integration and basic combinatorics. Consider the following integral:

$$\displaystyle\int_{0}^{2\pi} \! \exp{(2\cos \theta)} \, \mathrm{d} \theta.$$

No apparent tricks from ordinary calculus appear to be helpful, so we turn instead to complex variables. The limits of integration suggest a parametrization of the unit circle. Let $$C$$ denote the unit circle, oriented counterclockwise. We’d like to convert our original integral into a contour integral of the form $$\oint_C \! f(z) \, \mathrm{d} z$$ for an appropriate function $$f(z)$$. We can parametrize the contour by letting $$z = e^{i\theta}, 0 \le \theta \le 2\pi.$$ Then $$\mathrm{d} z = ie^{i\theta} \mathrm{d} \theta$$, so that $$\mathrm{d} \theta = \frac{1}{i} e^{-i\theta} \mathrm{d} z$$, or in other words, $$\mathrm{d} \theta = \frac{1}{i} z^{-1} \mathrm{d} z.$$ We also observe that, on our contour, $$z + z^{-1} = e^{i\theta} + e^{-i\theta} = (\cos \theta + i \sin \theta) + (\cos \theta – i \sin \theta) = 2 \cos \theta$$. Hence,

$$\displaystyle\int_{0}^{2\pi} \! \exp{(2\cos \theta)} \, \mathrm{d} \theta = \displaystyle\oint_C \! \exp{(z + z^{-1})}\frac{1}{i} z^{-1} \, \mathrm{d} z.$$

Within the region bounded by the contour, the integrand has exactly one singularity. More precisely, it has an essential singularity at $$z=0$$. Thus, by Cauchy’s residue theorem, the value of the integral is equal to $$2\pi i$$ times the integrand’s residue at $$z=0$$.

Let $$f(z) = z^{-1} \exp{(z + z^{-1})}$$. To find the residue of $$f$$ at $$0$$, we consider the Laurent series for $$f$$ centered at $$0$$. (A simple approach here is to write $$\exp{(z + z^{-1})} = \exp{(z)} \exp{(z^{-1})}$$, but we shall take a more roundabout path. The reader is invited to attempt the direct route.) Using the standard series for the exponential, we obtain

$$f(z) = z^{-1} \displaystyle\sum_{n=0}^{\infty} \frac{(z + z^{-1})^n} {n!}.$$

The residue of $$f$$ at $$z = 0$$ is defined as the coefficient of the $$z^{-1}$$ term in the Laurent series for $$f$$ centered at $$0$$. From the above expansion, we see that in order to determine this coefficient, it suffices to determine the coefficient of $$z^0$$ within the infinite sum.

Fix a nonnegative integer $$n$$ and consider the binomial $$(z + z^{-1})^n$$. We wish to determine the coefficient of $$z^0$$ in the expansion of this binomial. In order to obtain a $$z^0$$ term in the first place, we see that the terms $$z$$ and $$z^{-1}$$ must multiply together in pairs. Hence, if $$n$$ is odd, the coefficient of $$z^0$$ is $$0$$. Thus, we may focus our attention on the nonnegative even integers.

Consider the binomial $$(z + z^{-1})^{2n}$$. What is the coefficient of $$z^0$$ in its expansion? If we imagine $$2n$$ copies of the expression $$(z + z^{-1})$$ being multiplied together, the only way we obtain a nonzero $$z^0$$ term is when we choose the $$z$$ and $$z^{-1}$$ terms in pairs. There will be a total of $$n$$ pairs. Of the $$2n$$ copies of $$z$$, we can choose any $$n$$ of them, and then the remaining $$n$$ choices will have to be $$z^{-1}$$ terms. Hence, the coefficient of $$z^0$$ is $${{2n}\choose{n}}$$.

Since, in our original series, each term $$(z + z^{-1})^n$$ is divided by $$n !$$, we must divide $${{2n}\choose{n}}$$ by $$(2n)!$$. By definition,  $${{2n}\choose{n}} = \frac{(2n)!}{n! n!}$$, so  $${{2n}\choose{n}}$$ divided by $$(2n)!$$ is simply $$\frac{1}{n! n!}$$.

Thus, the desired residue is seen to be

$$\displaystyle\sum_{n=0}^{\infty} \frac{1}{n! n!}.$$

It follows by the residue theorem that the value of the integral we seek is equal to

$$\displaystyle 2\pi i \frac{1}{i} \sum_{n=0}^{\infty} \frac{1}{n! n!} = 2\pi \displaystyle\sum_{n=0}^{\infty} \frac{1}{n! n!}. \, \spadesuit$$

It turns out that the original integral is not entirely unmotivated, as it can also be written as $$2\pi I_{0}(2)$$, where $$I_{n}(z)$$ denotes the modified Bessel function of the first kind. This result can be seen from the integral formulas for these functions.