Modified Bessel Function Integral representation proof $K_{\nu}(z)=\frac{z^{\nu}}{2^{\nu+1}}\int_{0}^{\infty}t^{-\nu-1}e^{-t-z^{2}/4t}dt $ How do I proof the following integral representation for the Modified Bessel function of the second kind (Macdonald Function).
$K_{\nu}(z)=\frac{z^{\nu}}{2^{\nu+1}}\int_{0}^{\infty}t^{-\nu-1}e^{-t-z^{2}/4t}dt $
Thank you in advance.
 A: Here's a way to show that it works, without actually deriving it: one definition of $K_{\nu}$ (and often the useful one, given what we normally do with it) is that it is the unique solution of the modified Bessel equation,
$$ z^2 w''+zw'-(z^2+\nu^2)w=0, $$
such that
$$ w \sim e^{-z}\sqrt{\frac{\pi}{2z}} $$
as $z \to \infty$ (this exists, using Liouville-Green; it is unique, since examining the Wronskian shows that any other solution must blow up as $z\to \infty$).
You can show, using differentiation under the integral sign and integration by parts, that your integral satisfies the modified Bessel equation (it is probably easiest to set $e^u=2t/z$, and use the resulting form
$$ \frac{1}{2}\int_0^{\infty} e^{-z\cosh{u}} e^{-\nu u} \, du $$
to show this, since the $z$-content is then minimal). It remains to check the asymptotics. Using the form I just mentioned, it is in the form of a Laplace integral, with a unique maximum in the coefficient of $z$ in the exponent at $u=0$; hence applying Laplace's method, we find
$$ \frac{1}{2}\int_0^{\infty} e^{-z\cosh{u}} e^{-\nu u} \, du \sim \frac{1}{2} \int_0^{\infty} e^{-z(1+u^2/2)} 1 \, du = e^{-z}\frac{\sqrt{2\pi}}{2\sqrt{z}}, $$
and so it has the correct asymptotics. Hence it must be $K_{\nu}(z)$.
(The standard way for deriving this representation uses some trickery involving the $\Gamma$-function and the Hankel contour, along with several nasty changes of variable and contour integrals, so I think this proof, while nonconstructive, is nicer. You can see the full proof in Watson's book A Treatise on the Theory of Bessel Functions, on p. 181ff..)
