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