Integral representation of the modified Bessel functions of the second kind and asymptotic expansion - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:56:38Z http://mathoverflow.net/feeds/question/120420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120420/integral-representation-of-the-modified-bessel-functions-of-the-second-kind-and-a Integral representation of the modified Bessel functions of the second kind and asymptotic expansion Sasha 2013-01-31T14:45:13Z 2013-01-31T15:37:20Z <p>The modified Bessel function (Macdonald function) $K_\alpha(z)$ is known to have the following asymptotic expansion for large positive $z$: $$K_\alpha(z)=\sqrt{\frac{\pi}{2z}}e^{-z}\sum_{k=0}^\infty \frac{b_k(\alpha)}{z^ k}$$ where $b_1(\alpha)=1$, $b_2(\alpha)=\frac{\alpha^2-1^2}{1!8}$, $b_3(\alpha)=\frac{(\alpha^2-1^2)(\alpha^2-3^2)}{2!(8)^2}$ and so on. Is there any simple integral representation, for which it would be a perturbative expansion such that $$K_\alpha(z)=h(z) \int_C \exp\left(\frac{f(y)}{z}\right) g(y)^\alpha d\mu(y)$$ where $f(x)$, $g(x)$, $h(x)$ and $d \mu(x)$ are $\alpha$-independent? </p> http://mathoverflow.net/questions/120420/integral-representation-of-the-modified-bessel-functions-of-the-second-kind-and-a/120425#120425 Answer by Igor Khavkine for Integral representation of the modified Bessel functions of the second kind and asymptotic expansion Igor Khavkine 2013-01-31T15:37:20Z 2013-01-31T15:37:20Z <p>The <a href="http://dlmf.nist.gov/" rel="nofollow">DLMF</a> lists multiple <a href="http://dlmf.nist.gov/10.32.i" rel="nofollow">integral representations</a> of $K_\nu(z)$. Here's one that fits your bill:</p> <p><code> $\mathop{K_{{\nu}}}\nolimits\!\left(z\right)=\frac{\pi^{{\frac{1}{2}}}(\frac{1}{2}z)^{\nu}}{\mathop{\Gamma}\nolimits\!\left(\nu+\frac{1}{2}\right)}\int _{0}^{\infty}e^{{-z\mathop{\cosh}\nolimits t}}(\mathop{\sinh}\nolimits t)^{{2\nu}}dt .$</code></p> <p>For integer $\nu$, the contour could be extended by symmetry to all of $\mathbb{R}$.</p>