Dertivative of a Special Function with respect to Order - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:14:03Z http://mathoverflow.net/feeds/question/111910 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111910/dertivative-of-a-special-function-with-respect-to-order Dertivative of a Special Function with respect to Order Remy 2012-11-09T15:39:24Z 2012-11-09T17:51:54Z <p>The marcum Q-function is defined by $$Q_m(a,b) = \int^\infty_b x \left(\frac{x}{a}\right)^{m-1} \exp\left(-\frac{x^2+a^2}{2}\right) I_m\left(a x\right) \:\mathrm{d} x,$$</p> <p>where $m\in\mathbb{N}$ , $b\in\mathbb{R}^+$ , $a\in\mathbb{R}^+$ , and $I_m(.)$ is the $m$-th order modified Bessel function of the first type.</p> <p>Is it possible to get the derivative of the Q-function with respect to $m$? </p> http://mathoverflow.net/questions/111910/dertivative-of-a-special-function-with-respect-to-order/111920#111920 Answer by Emilio Pisanty for Dertivative of a Special Function with respect to Order Emilio Pisanty 2012-11-09T17:51:54Z 2012-11-09T17:51:54Z <p>For large $x$, if $a>0$, $I_m$ behaves asymptotically like $I_m(ax)\approx e^{ax}/\sqrt{ax}$. Therefore for large $x$ the integrand will look like $x^{m-1/2}e^{-(x-a)^2/2}$. This dies off fast enough that the improper integral converges uniformly in $m$ and you can differentiate inside the integral.</p>