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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..

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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: $$\sqrt{\pi}\,\left(I_0(\lambda|a+b|)\log I_0(\lambda|a+b|)\right)^{\frac{3}{2}}\left(\frac{\lambda ab\,I_1(\lambda|a+b|)}{2|a+b|}\left(1+\log I_0(\lambda |a+b|)\right)\right)^{-\frac{1}{2}}.$$

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  • $\begingroup$ Thanks for the answer. I am a bit lost after the Taylor series expression. It looks like you did not integrate term by term over the range of interest. Would you mind to explain how you got the final expression? Best regards.. @jack-daurizio $\endgroup$ – mkesal Feb 19 '15 at 17:44
  • $\begingroup$ @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 $a^{3/2}\sqrt{\frac{\pi}{b}}$. $\endgroup$ – Jack D'Aurizio Feb 19 '15 at 17:56
  • $\begingroup$ @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.. $\endgroup$ – mkesal Feb 19 '15 at 18:18

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