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 $h(\alpha)=t-(\frac{x}{\alpha})\cosh t$



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> 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 $\alpha\rightarrow\infty$ ?
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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.



Asymptotic expansion of modified Bessel function $K_\alpha$
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


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


