Integrate $\int_{0}^{2\pi}e^{-[a\cos(2\psi)+b\cos(\psi)]}d\psi$ and $\int_{0}^{2\pi}e^{-[a\cos(\phi+\psi)+b\cos(\phi-\psi)]}d\psi$

Question

I'm trying to determine the following integrals: $$\int_{0}^{2\pi}e^{-[a\cos(2\psi)+b\cos(\psi)]}d\psi$$ and $$\int_{0}^{2\pi}e^{-[a\cos(\phi+\psi)+b\cos(\phi-\psi)]}d\psi.$$ Similar integrals can be solved by means of the modified Bessel function of the first kind of order zero. Any help is appreciated!

Answer 1

The first integral can be expressed as a series of products of modified Bessel functions. The Jacobi-Anger expansion for the modified Bessel functions of the first kind reads \begin{equation} e^{z\cos\theta}=I_{0}\left(z\right)+2\sum_{k=1}^{\infty}I_{k}\left(z\right)% \cos\left(k\theta\right) \end{equation} then \begin{align} e^{-[a\cos(2\psi)+b\cos(\psi)]}&=\left( I_{0}\left(-a\right)+2\sum_{k=1}^{\infty}I_{k}\left(-a\right)\cos\left(2k\psi\right) \right)\left( I_{0}\left(-b\right)+2\sum_{m=1}^{\infty}I_{m}\left(-b\right)\cos\left(m\psi\right) \right)\\ &= I_{0}\left(a\right) I_{0}\left(b\right)+2 I_{0}\left(a\right)\sum_{m=1}^{\infty}(-1)^mI_{m}\left(b\right)\cos\left(m\psi\right)\\ &\hspace{2cm}+2 I_{0}\left(b\right)\sum_{k=1}^{\infty}(-1)^kI_{k}\left(a\right)\cos\left(2k\psi\right)\\ &\hspace{2cm}+4\sum_{k,m=1}^{\infty}(-1)^{k+m}I_{k}\left(a\right)I_{m}\left(b\right)\cos\left(2k\psi\right)\cos\left(m\psi\right) \end{align} After integration over $(0,2\pi)$, the second and third series vanish while, in the double sum, only the terms with $m=2k$ survive with \begin{equation} \int_0^{2\pi}\cos^22k\psi\,d\psi=\pi \end{equation} Then \begin{align} F_1&=\int_{0}^{2\pi}e^{-[a\cos(2\psi)+b\cos(\psi)]}\,d\psi\\ &=2\pi I_{0}\left(a\right) I_{0}\left(b\right)+4\pi\sum_{k=1}^\infty(-1)^kI_{k}\left(a\right)I_{2k}\left(b\right) \end{align} This result and many similar ones are obtained in the frame of the generalization of Bessel functions to several variables (see Gupta or the many papers by Dattoli, for example).

The second integral can be put into a closed form, by remarking that one can transform \begin{align} a\cos(\phi+\psi)+b\cos(\phi-\psi)&=(a+b)\cos\phi\cos\psi-(a-b)\sin\phi\sin\psi\\ &=\Delta\left(\frac{(a+b)\cos\phi}{\Delta} \cos\psi- \frac{(a-b)\sin\phi}{\Delta}\sin\psi\right) \end{align} where \begin{align} \Delta&=\sqrt{(a+b)^2\cos^2\phi+(a-b)^2\sin^2\phi }\\ &=\sqrt{a^2+b^2+2ab\cos2\phi} \end{align} Then defining an angle $0\le\theta<\pi$ by $\tan\theta=\frac{a-b}{a+b}\tan\phi$, the integral reads \begin{align} F_2&=\int_{0}^{2\pi}e^{-[a\cos(\phi+\psi)+b\cos(\phi-\psi)]}\,d\psi\\ &=\int_{0}^{2\pi}e^{-\Delta(\cos\theta\cos\phi-\sin\theta\cos\phi)}\,d\psi\\ &=\int_{0}^{2\pi}e^{-\Delta\cos(\psi+\theta)}\,d\psi\\ &=\int_{0}^{2\pi}e^{-\Delta\cos\psi}\,d\psi\\ =&2\int_{0}^{\pi}e^{-\Delta\cos\psi}\,d\psi \end{align} where the integration variable was shifted, as the domain of integration covers a period of the function and where we used the symmetry of the integrand. From the integral representation for the modified Bessel function of the first kind and of order $0$: \begin{align} I_{0}\left(\Delta\right)=\frac{1}{\pi}\int_{0}^{\pi}e^{\pm \Delta\cos\psi} \theta\,d\psi \end{align} we deduce \begin{equation} F_2=2\pi I_0\left(\sqrt{a^2+b^2+2ab\cos2\phi} \right) \end{equation}