# Asymptotic Expansion of an Integral involving Modified Bessel Functions

I do not have enough experience with the asymptotic expansion of integrals especially involving Bessel functions. I appreciate any feedback that you guys provide. Here is the problem. Let $a$ and $b$ be non-negative finite valued real numbers. Consider the following integral \begin{align} T(\lambda)=\int_{-\pi}^{\pi} I_0\left(\lambda\sqrt{a^2+b^2+2ab\cos(x)}\right)\,\log I_0\left(\lambda\sqrt{a^2+b^2+2ab\cos(x)}\right)\,\mathrm{d}x \end{align} As $\lambda\to\infty$, what would be the leading term of the asymptotic expansion of $T(\lambda)$? Here, $I_0(x)$ is the modified Bessel function of the first kind with order $0$. I attempted to use the Laplace integration method and ended up with \begin{align} T(\lambda)\sim \frac{e^{\lambda(a+b)}\log I_0(\lambda(a+b))}{\lambda \sqrt{ab}} \end{align} I have no clue how to verify this. Thanks a lot..

$$I_0(\lambda\sqrt{a^2+b^2+2ab\cos x})\log I_0(\lambda\sqrt{a^2+b^2+2ab\cos x}) \\= I_0(\lambda|a+b|)\log I_0(\lambda|a+b|)-\frac{\lambda ab\,I_1(\lambda|a+b|)}{2|a+b|}\left(1+\log I_0(\lambda |a+b|)\right)x^2$$
we have that the asymptotic behaviour of the integral is expected to be: $$\sqrt{\pi}\,\left(I_0(\lambda|a+b|)\log I_0(\lambda|a+b|)\right)^{\frac{3}{2}}\left(\frac{\lambda ab\,I_1(\lambda|a+b|)}{2|a+b|}\left(1+\log I_0(\lambda |a+b|)\right)\right)^{-\frac{1}{2}}.$$
• @mkesal: how did you apply Laplace method? Here we have a gaussian-like function whose behaviour in zero is $a-bx^2$, hence we approximate such a function (quite well, indeed) by $a e^{-\frac{b}{a}x^2}$ and integrate the last function over $\mathbb{R}$ to get $a^{3/2}\sqrt{\frac{\pi}{b}}$. – Jack D'Aurizio Feb 19 '15 at 17:56
• @jack-daurizio Actually, as $\lambda$ tends to infinity, this function converges to Gaussian density up to a scaling constant. You might have noticed that this indeed is a convolution of two von-Mises density functions over a finite interval and they are known to converge to Gaussian density as well. I replaced the first term with its first order asymptotic and the arguments of the bessels as $\hat{\lambda}\sqrt{1-c\sin^2(x/2)}$. Then the Laplace.. I got the idea of your method and it is quite neat. The main challenge would be to justify bell shape replacement rigorously. Best.. – mkesal Feb 19 '15 at 18:18