I am trying to find the following integral in the close form:
$\int\limits_0^{2\pi } {\exp \left( {ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta ) + c{r^3}\cos ({\psi _c} - \theta )} \right)} d\theta $
Where, $a,b,c,r$ is non-negative real number.
I was trying to use the form of zero-order modified Bessel identity that is, Assuming $a > 0$, $$ I_0(x) = \frac{1}{\pi}\int_0^\pi e^{x\cos\theta}d\theta $$ Where $I_0(x)$ is a modified Bessel function of the first kind.
My approach:
- Converting ${ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta ) + c{r^3}\cos ({\psi _c} - \theta )}=Kcos({\psi _K}-\theta)$. Then use the identity mentioned above. But I am not sure if it is possible to convert ${ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta ) + c{r^3}\cos ({\psi _c} - \theta )}=Kcos({\psi _K}-\theta)$?
- Is there any other approach that I might take? Any suggestion would help.
I shall try to make it completely general.
Changing notations, the agument of the exponential function is $$S=\sum_{n=1}^p a_n r^n \cos (\psi_n-\theta) $$ Expanding the cosines, we have $$S=\Big[\sum_{n=1}^p a_n r^n \cos (\psi_n)\Big]\cos (\theta)+\Big[\sum_{n=1}^p a_n r^n \sin (\psi_n)\Big]\sin (\theta)$$ that is to say $$S=A_c \,\cos (\theta)+A_s \,\sin (\theta)$$ and then the problem is $$\int_0^{2\pi} \exp\Big[A_c \,\cos (\theta)+A_s \,\sin (\theta)\Big]\,d\theta=2 \pi I_0\left(\sqrt{A_c^2+A_s^2}\right)$$