2
$\begingroup$

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 $m$?

$\endgroup$
2
  • 1
    $\begingroup$ In fact, you just need the derivative w.r.t. $m$ of $(x/a)^m I_m(ax)$, right ? $\endgroup$
    – Lierre
    Nov 9 '12 at 15:44
  • $\begingroup$ 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 $m$ $\endgroup$
    – Remy
    Nov 9 '12 at 17:17
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.