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.

  • $\begingroup$ 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! $\endgroup$
    – Lucia
    Feb 25, 2014 at 20:22

1 Answer 1


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


  • $\begingroup$ Thanks for your answer, I did not expect such a short one. My remaining question is how did you find this specific form of $x_0$ $\endgroup$
    – Montaigne
    Feb 25, 2014 at 10:41
  • $\begingroup$ $x_0$ is the curvature of the exponent around the saddle point at $t=t_0$. $\endgroup$ Feb 25, 2014 at 11:18

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