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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 $\mathbf{M}$.

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    $\begingroup$ 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 $\endgroup$
    – tired
    Aug 30, 2016 at 10:14
  • $\begingroup$ multiply everything above by $2\pi$ ^^ $\endgroup$
    – tired
    Aug 30, 2016 at 10:23
  • $\begingroup$ Out of curiosity, where does this integral come from? $\endgroup$ Aug 30, 2016 at 10:36
  • $\begingroup$ 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. $\endgroup$
    – bkocsis
    Aug 31, 2016 at 16:30

1 Answer 1

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The following is just the sketch of an idea. With some luck, somebody will be able to complete it (or show that it is wrong, of course!)

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.

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  • $\begingroup$ the exponetial is way to strong to allow the second term go to zero $\endgroup$
    – tired
    Aug 30, 2016 at 10:37
  • $\begingroup$ @tired Mmh, yeah, you're almost certainly correct. $\endgroup$ Aug 30, 2016 at 10:40

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