Closed-form expression for integral involving exponential of cosines?

Question

I'm looking for a closed-form expression for the following integral: $$ \int_0^{2\,\pi} \exp\{a_1\,\cos(\theta_1-\mu) + a_2\,\cos(\theta_2 - \mu)\}\,d\mu $$ where $a_1,a_2 > 0$ and $0 \le \theta_1, \theta_2 < 2\,\pi$.

Answer 1

We wish to evaluate

$$I(a_1,a_2,\theta_1,\theta_2):=\int_0^{2\pi}\exp\left(a_1\cos(\theta_1-\mu)+a_2\cos(\theta_2-\mu)\right)d\mu \tag1$$

Note that

$$\cos(x-y) = \sin x\sin y + \cos x\cos y \tag2$$

Letting $s_k:=\sin\theta_k$, $c_k:=\cos\theta_k$ for $k\in\{1,2\}$ and using $(2)$, $(1)$ becomes

\begin{align} I(a_1,a_2,s_1,s_2)=\int_0^{2\pi}\exp\left(a_1(s_1\sin\mu+c_1\cos\mu)+a_2(s_2\sin\mu+c_2\cos\mu)\right)d\mu \tag3 \end{align}

Letting $S:=a_1s_1+a_2s_2$ and $C=a_1c_1+a_2c_2$ and regrouping terms, $(3)$ is equivalent to

\begin{align} I(C,S)&=\int_0^{2\pi}\exp\left(S\sin\mu+C\cos\mu\right)d\mu\\ &=\int_0^{2\pi}\exp\left(\frac{S}{2i}\left(e^{i\mu}-e^{-i\mu}\right)+\frac{C}{2}\left(e^{i\mu}+e^{-i\mu}\right)\right)d\mu\\ &=\int_0^{2\pi}\exp\left(\left(\frac{C}{2}+\frac{S}{2i}\right)e^{i\mu}+\left(\frac{C}{2}-\frac{S}{2i}\right)e^{-i\mu}\right)d\mu \tag4 \end{align}

We define $\gamma:=\frac{C}{2}+i\frac{S}{2}$. With this, we finally simplify the integral in $(4)$ to

\begin{align} I(\gamma)&=\int_0^{2\pi}\exp\left(\gamma^*e^{i\mu}+\gamma e^{-i\mu}\right)d\mu \tag5 \end{align}

where $\gamma^*$ denotes the complex conjugate of $\gamma$.

We now transform this into a contour integral where the contour $\Gamma$ we consider is the unit circle, traversed in the counterclockwise direction. On the unit circle, $z=e^{i\mu}$, thus $e^{-i\mu}=\frac{1}{z}$ and $d\mu = \frac{dz}{iz}$.

With this, $(5)$ becomes

\begin{align} I(\gamma)&=\frac{1}{i}\oint_\Gamma e^{\gamma^*z}e^{\frac{\gamma}{z}}\frac{dz}{z}\\ &=\frac{1}{i}\oint_\Gamma \frac{e^{\gamma^*z}}{z}\sum_{k=0}^\infty\frac{\gamma^k}{k!z^k}dz \tag6\\ &=\frac{1}{i}\sum_{k=0}^\infty\frac{\gamma^k}{k!}\oint_\Gamma\frac{e^{\gamma^*z}}{z^{k+1}}\\ &=2\pi\sum_{k=0}^\infty\frac{\gamma^k}{k!}\text{Res}\left(\frac{e^{\gamma^*z}}{z^{k+1}}, 0\right) \tag7\\ &=2\pi\sum_{k=0}^\infty\frac{\gamma^k}{k!}\frac{1}{k!}\lim\limits_{z\rightarrow 0}\frac{d^k}{dz^k}e^{\gamma^*z}\\ &=2\pi\sum_{k=0}^\infty\frac{|\gamma|^{2k}}{(k!)^2}\tag8\\ \end{align}

where $(6)$ uses the series expansion of the exponential function, $(7)$ uses the Residue Theorem and $(8)$ uses that $|z|^2=zz^*$.

Reversing our substitutions to get the answer in terms of the original variables, we have

\begin{align} I(a_1,a_2,\theta_1,\theta_2)=2\pi\sum_{k=0}^\infty\frac{\left((a_1\cos\theta_1+a_2\cos\theta_2)^2+(a_1\sin\theta_1+a_2\sin\theta_2)^2\right)^k}{4^k(k!)^2} \tag9 \end{align}

$(9)$ is a series representation of the Modified Bessel Function of the First Kind $I_0$.

Thus, your integral is given by

\begin{align} I(a_1,a_2,\theta_1,\theta_2)=2\pi I_0\left(\sqrt{(a_1\cos\theta_1+a_2\cos\theta_2)^2+(a_1\sin\theta_1+a_2\sin\theta_2)^2}\right) \end{align}