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The marcum Q-function is defined by $$ Q_m(a,b) = \int^\infty_b x \left(\frac{x}{a}\right)^{m-1} \exp\left(-\frac{x^2+a^2}{2}\right) I_m\left(a x\right) \:\mathrm{d} x,$$

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.

Is it possible to get the derivative of the Q-function with respect to $m$?

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In fact, you just need the derivative w.r.t. $m$ of $(x/a)^m I_m(ax)$, right ? –  Lierre Nov 9 '12 at 15:44
Thanks Lierre, you are right, I was just not sure about that because it is a derivative under the integral sign...but you are right... it the derivative of $(x/a)^m I_m(ax)$ w.r.t $m$ –  Remy Nov 9 '12 at 17:17
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1 Answer 1

up vote 2 down vote accepted

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.

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