# Evaluating $\frac{1}{\pi} \int_{0}^{\pi} e^{2\cos{\theta}} d\theta$

I ran into this integral when computing the volume of a family of polytopes and I'm not sure how to evaluate it analytically (I know Wolframalpha says 2.27...). Any ideas? I tried using complex analysis (Cauchy Integral Formula, Residue Theorem, etc.) but nothing seemed applicable.

$$\frac{1}{\pi} \int_{0}^{\pi} e^{2\cos{\theta}} d\theta$$

• Mathematica says it is equal to $I_0(2)$, where $I_0$ is the modified Bessel function of $0$th order. It is an easy consequence of the integral representation of this function (see formula (4) here). Commented Jun 24, 2013 at 19:06
• According to Gradshteyn and Ryzhik, the integral evaluates to $I_0(2)$, where $I_0$ is the modified Bessel function of the first kind. Commented Jun 24, 2013 at 19:06
• @Samuel Reid :$$\frac{1}{\pi} \int_{0}^{\pi} e^{2\cos{\theta}} d\theta=\frac{-1}{\pi}\int_{-1}^{0}\frac1ze^{z+\overline z}dz$$
– M.H
Commented Jun 24, 2013 at 19:19

$$I=\frac{1}{\pi}\int_0^\pi e^{2\cos\theta}\,d\theta=\frac{1}{\pi}\sum_{k=0}^\infty\frac{2^k}{k!}A_k,$$ with $$A_k=\int_0^\pi\cos^k\theta\,d\theta \quad \forall k \ge 0.$$ We have $$A_0=\int_0^\pi\,d\theta=\pi.$$ For every $k \ge 1$, if we set $\varphi=\pi-\theta$, then $$\int_{\pi/2}^\pi\cos^k\theta\,d\theta=\int_0^{\pi/2}\cos^k(\pi-\varphi)\,d\varphi=(-1)^k\int_0^{\pi/2}\cos^k\varphi\,d\varphi.$$ Theorefore, for every $k \ge 1$ we have $$A_k=\int_0^{\pi/2}\cos^k\theta\,d\theta+\int_{\pi/2}^\pi\cos^k\theta\,d\theta=[1+(-1)^k]\int_0^{\pi/2}\cos^k\theta\,d\theta.$$ We deduce that $A_{2k+1}=0$ for all $k \ge 0$. For every $k\ge 1$ we have \begin{eqnarray} B_k&:=&A_{2k}=2\int_0^{\pi/2}\cos^{2k}\theta\,d\theta=2\int_0^{\pi/2}(\sin\theta)'\cos^{2k-1}\theta\,d\theta\\ &=&2(2k-1)\int_0^{\pi/2}\sin^2\theta\cos^{2k-2}\theta\,d\theta=2(2k-1)\int_0^{\pi/2}(1-\cos^2\theta)\cos^{2k-2}\theta\,d\theta\\ &=&(2k-1)(B_{k-1}-B_k), \end{eqnarray} i.e. $$B_k=\frac{2k-1}{2k}B_{k-1} \quad \forall k \ge 1.$$ Thus $$B_k=\frac{(2k-1)\cdot(2k-3)\ldots3\cdot1}{(2k)\cdot(2k-2)\ldots4\cdot2}B_0=\frac{(2k)!}{[(2k)(2k-2)\ldots4\cdot2]^2}B_0=\frac{(2k)!}{2^{2k}(k!)^2}\pi.$$ The given integral is then $$I=\frac{1}{\pi}\sum_{k=0}^\infty\frac{2^{2k}}{(2k)!}A_{2k}=\frac{1}{\pi}\sum_{k=0}^\infty\frac{2^{2k}}{(2k)!}\cdot\frac{(2k)!}{2^{2k}(k!)^2}\pi=\sum_{k=0}^\infty\frac{1}{(k!)^2}.$$

• There is an easier way to obtain this. Using parity, extend the integration to interval $(-\pi,\pi)$, then expand the exponential in Taylor series (as you do), and then expand $\cos^n\theta$ using binomial theorem. All the exponentials $e^{ik\theta}$ with $k\neq0$ will give zero integrals (the new bounds are crucial for this), so each $\cos^n\theta$ will produce only one non-zero term which more or less coincides with your $A_{2n}$. But +1 anyway. Commented Jun 24, 2013 at 20:25
• @Mercy: Thanks for this solution, I will combine O.L.'s idea with your solution here to figure it out! Commented Jun 24, 2013 at 23:53

Let me post my comment as an answer since there is not much more to say here.

Mathematica says that this integral is equal to $I_0(2)$, where $I_0$ denotes the modified Bessel function of $0$th order. It is an easy consequence of its standard integral representation (see formula (4) here).

Bessel functions do not reduce to simpler expressions at integer values of the argument (except $0$) nor for order $0$, which means that this expression cannot be simplified further.

• I was already aware of the representation in terms of Bessel functions, I was looking for an explicit solution as given in the other answer. Commented Jun 24, 2013 at 23:53