What is the integral $\int_{-\pi}^{\pi} \frac{1}{2\pi}\exp{(z_1 \cos\theta + z_2 \sin\theta)}\, d\theta$?

Question

What is the integral $\int_{-\pi}^{\pi} \frac{1}{2\pi}\exp{(z_1 \cos\theta + z_2 \sin\theta)}\, d\theta$? When $z_{1,2} \in \mathbb{R}$ then we get the modified Bessel function $I_0(\sqrt{z_1^2+z_2^2})$. What would be the solution for complex $z_i$?

Answer 1

For complex $z_1,z_2$, one has $$\begin{eqnarray} I(z_1,z_2)&=&\int_{-\pi}^{\pi} \frac{1}{2\pi}\exp{(z_1 \cos\theta + z_2 \sin\theta)}\, d\theta\\ &\overset{z=e^{i\theta}}=& \frac{1}{2\pi}\int_{|z|=1}\exp{\left(\frac12z_1(z+z^{-1}) + \frac1{2i}z_2(z-z^{-1})\right)}\, \frac{dz}{iz}\\ &=& \frac{1}{2\pi i}\int_{|z|=1}\frac1z\exp{\left(\frac12(z_1-z_2i)z+\frac1{2}(z_1+z_2i)z^{-1}\right)}\, dz\\ &=& \frac{1}{2\pi i}\int_{|z|=1}\frac1z\sum_{n=0}^\infty\frac1{n!}\left(\frac12(z_1-z_2i)z+\frac1{2}(z_1+z_2i)z^{-1}\right)^n\, dz\\ &=&\sum_{n=0}^\infty\frac{1}{(2n)!}\binom{2n}{n}\left(\frac12(z_1-z_2i)\cdot\frac1{2}(z_1+z_2i)\right)^{n}\\ &=&\sum_{n=0}^\infty\frac{1}{2^{2n}(n!)^2}(z_1^2+z_2^2)^n. \end{eqnarray}$$ Let $z^2=z_1^2+z_2^2$ and choose proper $z$. Then $I(z_1,z_2)=I_0(z)$.

Answer 2

For complex \begin{equation}( z1, z2 )\end{equation}, one has \begin{equation} I(z_1,z_2) = \int_{-\pi}^{\pi} \frac{1}{2\pi} \exp{(z_1 \cos\theta + z_2 \sin\theta)}\, d\theta \end{equation}

This integral can be recognized as related to the generating function of the Bessel functions. We can write:

\begin{align} I(z_1,z_2) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \exp\left(\Re\{z_1 e^{i\theta}\}\right) d\theta \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \exp\left(\frac{z_1}{2}(e^{i\theta} + e^{-i\theta}) + \frac{z_2}{2i}(e^{i\theta} - e^{-i\theta})\right) d\theta \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \exp\left(\frac{z_1}{2}(2\cos\theta) + \frac{z_2}{2i}(2i\sin\theta)\right) d\theta \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \exp\left(\frac{z_1}{2}(2\cos\theta) + \frac{z_2}{2}(2\sin\theta)\right) d\theta \\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \exp\left(\frac{z_1 + iz_2}{2} e^{i\theta} + \frac{z_1 - iz_2}{2} e^{-i\theta}\right) d\theta. \end{align}

Letting \begin{equation}(z^2 = z_1^2 + z_2^2)\end{equation}, and recognizing that the integral represents the zeroth-order modified Bessel function of the first kind for the complex argument (z), we get:

\begin{equation} I(z_1,z_2) = I_0(\sqrt{z_1^2 + z_2^2}) = I_0(z). \end{equation}