Does anyone know if the following integral can be written in a closed form expression and how? Thank you all;
$$I = \int_{\phi=0}^{2\pi} \sin\phi e^{\alpha\cos\phi+\beta\sin\phi} d\phi, \alpha, \beta \in\mathbb{R}^{+}$$
I am suspecting that the Modified Bessel function is involved, but I got stuck.
My attempt:
$$ I =\int_{0}^{2\pi} \frac{d}{d\beta} e^{\alpha\cos\phi+\beta\sin\phi} d\phi $$
$$ = \frac{d}{d\beta} \int_{0}^{2\pi} e^{\alpha\cos\phi+\beta\sin\phi} d\phi$$
$$= \frac{d}{d\beta} \int_{0}^{2\pi} e^{\sqrt{\alpha^{2}+\beta^{2}}\cos(\phi-\phi_{0})} d\phi$$
$$ = \frac{d}{d\beta} \int_{t=0}^{2\pi} e^{\sqrt{\alpha^{2}+\beta^{2}}\cos(t)} dt $$
$$= \frac{d}{d\beta} I_{0}\left(\sqrt{\alpha^{2}+\beta^{2}}\right)$$
$$ =2\pi\frac{\beta}{\sqrt{\alpha^{2}+\beta^{2}}} I_{1}\left(\sqrt{\alpha^{2}+\beta^{2}}\right), $$
Where $ \phi_{0}= \arccos \left( \dfrac{\alpha}{\sqrt{\alpha^{2}+\beta^{2}}} \right)$ and in the last step I used the correspondent property for the derivative of the Modified Bessel function.
I am not sure if my attempt is correct, and if I have to explain something for the exchange of the derivative with the integral.
Best Answer
Given $\alpha,\beta$ let $\rho=\sqrt{\alpha^2+\beta^2}$ and choose $\theta$ such that $\alpha = \rho \cos\theta,\beta = \rho \sin\theta$. Then the exponent becomes $$\rho \cos\theta \cos \phi +\rho \sin\theta \sin \phi=\rho\cos(\phi-\theta)$$ and thus the integral becomes $$I=\int_0^{2\pi} \sin\phi~ e^{\rho \cos(\phi-\theta)}d\phi$$ Shifting $\phi\to \phi+\theta$ and using periodicity, we obtain
\begin{align} I&=\int_0^{2\pi}\sin(\phi+\theta)e^{\rho \cos \phi}\,d\phi\\ &=\cos\theta\int_0^{2\pi}\sin \phi~ e^{\rho \cos \phi}\,d\phi+\sin\theta\int_0^{2\pi}\cos \phi~ e^{\rho \cos \phi}\,d\phi \end{align} The first integral has an explicit antiderivative and vanishes:
$$\int_0^{2\pi}\sin \phi~e^{\rho \cos\phi}~d\phi=\left[-\frac{1}{\rho}e^{\rho \cos\phi}\right]_{0}^{2\pi}=0$$ That leaves the second integral, which corresponds to a standard integral representation of the modified Bessel function of the first kind $I_1(\rho)$. Hence $$I=\sin \theta \cdot 2\pi I_1(\rho)=\frac{2\pi \beta}{\sqrt{\alpha^2+\beta^2}}I_1\left(\sqrt{\alpha^2+\beta^2}\right)$$