Asymptotic Expansion of an Integral involving Modified Bessel Functions

Question

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..

Answer 1

The integrand function is very Gaussian-shaped. If we compute the first three terms of the Taylor series in zero:

$$ I_0(\lambda\sqrt{a^2+b^2+2ab\cos x})\log I_0(\lambda\sqrt{a^2+b^2+2ab\cos x}) \\= I_0(\lambda|a+b|)\log I_0(\lambda|a+b|)-\frac{\lambda ab\,I_1(\lambda|a+b|)}{2|a+b|}\left(1+\log I_0(\lambda |a+b|)\right)x^2$$

we have that the asymptotic behaviour of the integral is expected to be: $$\sqrt{\pi}\,\left(I_0(\lambda|a+b|)\log I_0(\lambda|a+b|)\right)^{\frac{3}{2}}\left(\frac{\lambda ab\,I_1(\lambda|a+b|)}{2|a+b|}\left(1+\log I_0(\lambda |a+b|)\right)\right)^{-\frac{1}{2}}.$$