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).

  • $\begingroup$ Fantastic work! $\endgroup$
    – K.defaoite
    Sep 1 at 20:59
  • $\begingroup$ Thanks @K.defaoite It was fun! $\endgroup$
    – Paul Enta
    Sep 1 at 21:03
  • 1
    $\begingroup$ Thanks! The solution is great! $\endgroup$
    – Rostam22
    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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.