# Asymptotics of Gaussian integral over the unit sphere

I would like to evaluate the integral asymptotically over the unit sphere surface $$Z =\int e^{a \cos^2 \theta + b \sin^2\theta\cos2\phi + c\cos\theta} d\Omega = \int\limits_{0}^{\pi}\int\limits_{0}^{2\pi} e^{a \cos^2 \theta + b \sin^2\theta\cos2\phi + c\cos\theta} \sin\theta d\phi d\theta$$ for $a\rightarrow \pm\infty$ and $b\rightarrow\infty$, if $c$ is set implictly by the constraint $$L = \frac{\partial \ln Z}{\partial c}$$ for a fixed $0<L<1$. With the variables change $s=\cos\theta$, \begin{align} Z &= 2\pi\int_{-1}^1 e^{a s^2+c s}\, I_0[b (1-s^2)] ds\,, \\ L &= \frac{\int_{-1}^1 s\, e^{a s^2+c s} I_0[b (1-s^2)]\, ds }{\int_{-1}^1 e^{a s^2+c s} I_0[b (1-s^2)]\, ds }\,. \end{align}

where $I_0(x)$ is the Bessel function of the first kind. (Alternatively, the $\theta$ integral may be evaluated for any $\phi$ using the error function.)

Is there a general method to look for the $a\rightarrow \pm\infty$ asymptotics of this integral?

Note that this is the integral of a Gaussian on the sphere in the sense that $$Z_1=\int_{\|\mathbf{x}\|=1} e^{-{\mathbf{x}\cdot \mathbf{M x} + \mathbf{v}\cdot\mathbf{x}}} dS$$ where the integration is over the unit sphere's surface, $\mathbf{M}$ is a $3\times 3$ symmetric traceless matrix, and $\mathbf{v}$ is a vector parallel to one of the eigenvectors. The integral in spherical coordinates $\mathbf{x}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ and $dS = \sin\theta d\theta d\phi$ aligned with the eigenvectors, reduces to the integral given in the question as $Z_1=e^{-a/3} Z$, where $a=\frac32\lambda_3$, $b=\frac12(\lambda_1-\lambda_2)$, $\mathbf{v}=c\, \mathbf{u}_3$, and $\lambda_{1,2,3}$ and $\mathbf{u}_{1,2,3}$ are the three eigenvalues and normalized eigenvectors of $\mathbf{M}$.

• i claim that the asymptotics to leading order are given by $\sqrt{\pi}\frac{e^{a+c}}{4(a+c/2)^{3/2}}+\sqrt{\pi}\frac{e^{a-c}}{4(a-c/2)^{3/2}}$ which i found by a very similiar way then the answer to your last question concerning similar integrals. but there are some things i have to made rigorus so it will take some time to write down the answer Aug 30, 2016 at 10:14
• multiply everything above by $2\pi$ ^^ Aug 30, 2016 at 10:23
• Out of curiosity, where does this integral come from? Aug 30, 2016 at 10:36
• It comes from statistical mechanics. $Z$ is the partition function en.wikipedia.org/wiki/… of a system of stars orbiting around a supermassive black hole at the center of a galaxy, and $L$ is the total angular momentum of the system. We are looking for the equilibrium distribution of the angular momentum vector directions of stellar orbits. The $a\rightarrow \pm \infty$ limit is the ground state'' of the system, where the objects orbit in a thin disk. Aug 31, 2016 at 16:30

Differentiate with respect to $c$ to obtain \begin{align} \frac{\partial Z}{\partial c} = & \int_0^\pi\int_0^{2\pi}\left(e^{a\cos^2\theta}\cos\theta\sin\theta\right)e^{b\sin^2\theta\cos2\phi+c\cos\theta}d\phi d\theta\\ = & \int_0^\pi\int_0^{2\pi}\left(-\frac{1}{2a}\frac{\partial e^{a\cos^2\theta}}{\partial\theta}\right)e^{b\sin^2\theta\cos2\phi+c\cos\theta}d\phi d\theta. \end{align} Integrating by parts, we obtain \begin{align} \frac{\partial Z}{\partial c} = &\left[-\frac{1}{2a}\int_0^{2\pi}e^{a\cos^2\theta+b\sin^2\theta\cos2\phi+c\cos\theta}d\phi\right]_0^\pi\\ & +\frac{1}{2a}\int_0^\pi\int_0^{2\pi}e^{a\cos^2\theta+b\sin^2\theta\cos2\phi+c\cos\theta}\left(2b\sin\theta\cos\theta\cos2\phi - c\sin\theta\right)d\phi d\theta. \end{align} Now hopefully the term on the second line will go to zero as $a\to\infty$, and the first term can be evaluated using Bessel functions and the error function.
Edit: As remarked by @tired, the second term will not go to zero, as the exponential is increasing too fast as $a\to\infty$. I will leave this here anyway, so if someone wants to try something passing from partial integration as I did above they won't have to do the computation over again.