Asymptotic expansion of modified Bessel function $K_\alpha$

An integral representation for the Bessel function $K_\alpha$ for real $x>0$ is given by $$K_\alpha(x)=\frac{1}{2}\int_{-\infty}^{\infty}e^{\alpha h(t)}dt$$ where $h(\alpha)=t-(\frac{x}{\alpha})\cosh t$

Do you know any reference where its shown that using some generalization of Laplace method $$K_\alpha(x)\sim(\frac{\pi}{2\alpha})^{\frac{1}{2}}(\frac{2\alpha}{ex})^\alpha$$ as $\alpha\rightarrow\infty$ ?

I have difficulties doing the calculation myself, first we cann see that the integrand has a maximum at $$t=\sinh^{-1}(\alpha/x)\sim\log(2\alpha/x)$$ which rises the idea using the substitution $t=\log(2\alpha/x)+c$ but I it does not help me.

• A good source for such asymptotics is Higher Transcendental Functions, Volume 2" part of the Bateman project edited by A. Erdelyi (McGraw Hill 1953). You'll find this question discussed on pages 24--28, and in general the book should contain all you want to know about Bessel functions! – Lucia Feb 25 '14 at 20:22

abbreviate $t_0={\rm arsinh}(\alpha/x),\;\;x_0=x\sqrt{1+\alpha^2/x^2}>0$

expand the exponent around the saddle point, to second order: $$\alpha[t-(x/\alpha)\cosh t]=-x_0+\alpha t_0-\tfrac{1}{2}x_0(t-t_0)^2+{\rm order}(t-t_0)^3$$

carry out the Gaussian integration:

$$\int_{-\infty}^{\infty}\tfrac{1}{2}\exp\left(-x_0+\alpha t_0-\tfrac{1}{2}x_0(t-t_0)^2\right)dt=\sqrt{\frac{\pi}{2x_0}}\exp(-x_0+\alpha t_0)$$

take the limit $\alpha\rightarrow\infty$ and you're done: $x_0\rightarrow\alpha$, $t_0\rightarrow\ln(2\alpha/x)$, so

$$K_\alpha(x)\rightarrow\sqrt{\frac{\pi}{2\alpha}}e^{-\alpha}(2\alpha/x)^\alpha$$

• Thanks for your answer, I did not expect such a short one. My remaining question is how did you find this specific form of $x_0$ – Montaigne Feb 25 '14 at 10:41
• $x_0$ is the curvature of the exponent around the saddle point at $t=t_0$. – Carlo Beenakker Feb 25 '14 at 11:18