# Integral Involving Bessel function and trigonometric function

Consider the following integral

$$\begin{equation} \int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, I_0(B \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi \label{1} \end{equation}$$

where $$A$$ and $$B$$ are two constants and $$I_0(.)$$ is the modified Bessel function of the first kind.

To evaluate the integral, I tried to write the expansion of $$I_0$$ which (by Methemtica) results in $$\sum_{n=0}^{\infty} \frac{1}{4^n} \frac{1}{(n!)^2} \int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, (B \sin \phi \sin \theta)^{2n} \sin \theta \, d\theta \, d\phi$$ $$= \sum_{n=0}^{\infty} \frac{1}{4^n} \frac{1}{(n!)^2} \sqrt{\pi} \, \Gamma(n+1) \int_0^{2 \pi} \operatorname{HGR}_{01}\left(n+\frac{3}{2}, \frac{1}{4} A^2 \cos^2 \phi\right) \, (B^2 \sin^2 \phi)^{n} \sin \theta \, d\phi$$ where $$\operatorname{HGR}_{01}$$ is the regularized confluent hypergeometric function.

Again, by expanding this function I found (up to some constants) $$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} A'^m B'^n \frac{\Gamma(n+\frac{1}{2}) \Gamma(m+\frac{1}{2})\Gamma(m + n + \frac{3}{2})}{\Gamma(n+1) \Gamma(m+1)\Gamma(m+n+1)}$$ where $$A',B'$$ are (properly) rescaled version of $$A,B$$.

I couldn't go further, and stucked with this double sum.

I appreciate if anyone can provide me some insights on this sum, or any hints on the original integral .

In this paper by VNP Anghel, a result obtained by Glasser is generalized (eq. 32): $$\begin{equation} I_{\frac{m+1}{2}}(b)=\left( \frac{b}{2\pi \sin^m\alpha}\right)^{1/2} \int_0^\pi \exp(b\cos\alpha\cos\theta)I_{\frac m2}(b\sin\alpha\sin\theta)(\sin\theta)^{\frac{m+2}{2}}\,d\theta \tag{1}\label{eq1} \end{equation}$$ Then, in the case $$A=B=b$$, by choosing $$m=0$$, we have directly \begin{align} J(b,b)&=\int_0^{2 \pi} \int_0^{\pi} e^{b \cos \phi \cos \theta} \, I_0(b \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi\\ &=\int_0^{2 \pi} \, d\phi \left( \frac{2\pi}{b} \right)^{1/2}I_{1/2}(b)\\ &=4\pi\frac{\sinh b}{b} \end{align}

The result \eqref{eq1} may also be used in the case $$A\ne B$$ to express the integral as a series. We remark first that $$A$$ can be chosen to be positive (parity of the integral wrt $$A$$ is clear from the substitution $$\phi\to \pi-\phi$$ in the integral). By the multiplication theorem, $$\begin{equation} I_{\nu}\left(\lambda z\right)=\lambda^{\nu}\sum_{k=0}^{\infty} \frac{(\lambda^{2}-1)^{k}(\frac{1}{2}z)^{k}}{k!}I_{\nu+ k}\left(z \right) \end{equation}$$ with $$\nu=0,\lambda=B/A,z=A\sin\phi\sin\theta$$, one can express $$\begin{equation} I_0\left(B\sin\phi\sin\theta\right)=\sum_{k=0}^{\infty} \frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}\sin^k\phi\sin^k\theta \,I_{ k}\left(A\sin\phi\sin\theta\right) \end{equation}$$ and thus, by interverting integral and summation, \begin{align} J(A,B)&=\int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, I_0(B \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi\\ &=\sum_{k=0}^{\infty}\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}\int_0^{2 \pi}\sin^k\phi\,d\phi \int_0^{\pi} e^{A \cos \phi \cos \theta}I_{ k}\left(A\sin\phi\sin\theta\right) \sin^{k+1}\theta \, d\theta \end{align} With $$m=2k$$ in eq. \eqref{eq1}, it may be expressed as \begin{align} J(A,B)&=\sum_{k=0}^{\infty}\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}\int_0^{2 \pi}\left( \frac{2\pi\sin^{2k}\phi}{A} \right)^{1/2}\sin^k\phi \,I_{k+1/2}(A)\,d\phi\\ &=\sqrt{\frac{2\pi}{A}}\sum_{k=0}^{\infty}\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}I_{k+1/2}(A)\int_0^{2 \pi}\sin^{2k}\phi\,d\phi\\ &=\frac{(2\pi)^{3/2}}{\sqrt{A}}\sum_{k=0}^{\infty}\frac{(2k)!}{2^{3k}(k!)^3}\frac{\left( B^2-A^2 \right)^k}{A^k}I_{k+1/2}(A) \end{align} Numerical experiments show that this series converges very quickly, which is related to the asymptotic expansion of the modified Bessel function for large orders (DLMF).

• Fantastic work! Sep 1 at 20:59
• Thanks @K.defaoite It was fun! Sep 1 at 21:03
• Thanks! The solution is great! Sep 2 at 9:18

For the inner sum, another gaussian hypergeometric function.

$$\sum_{m=0}^{\infty} A^m B^n \frac{\Gamma(n+\frac{1}{2}) \Gamma(m+\frac{1}{2})\Gamma(m + n + \frac{3}{2})}{\Gamma(n+1) \Gamma(m+1)\Gamma(m+n+1)}=$$ $$\sqrt{\pi }\,\frac{ \Gamma \left(n+\frac{1}{2}\right) \Gamma \left(n+\frac{3}{2}\right)}{\Gamma (n+1)^2}\, _2F_1\left(\frac{1}{2},n+\frac{3}{2};n+1;A\right)\,B^n$$

Very little hope for the outer sum.