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What is the derivative of $Q_m\left(\frac{\alpha}{x^a},\frac{\beta}{x^b}\right)$ with respect to $x$, i.e,

$$\frac{\partial}{\partial x}Q_m\left(\frac{\alpha}{x^a},\frac{\beta}{x^b}\right), \quad \text{ where } \alpha,\beta,a,b \in \mathbb{R} \text{ and } m\in \mathbb{N} $$

where the $Q_m(.,.)$ is a special function( http://en.wikipedia.org/wiki/Marcum_Q-function ) , and we have

$$Q_m\left(\frac{\alpha}{x^a},\frac{\beta}{x^b}\right)= \left(\frac{x^a}{\alpha}\right)^{m-1} \int^\infty_{\frac{\beta}{x^b}} \: t^m \exp\left(-\frac{t^2}{2}-\frac{\alpha^2}{2x^{2a}}\right) I_{m-1}\left(\frac{\alpha}{x^a}t\right) \:\mathrm{d}t$$

where $I_m(.)$ is the modified bessel function of the first kind with order $m $. Please note that the lower limit of the integral is $\displaystyle\frac{\beta}{x^b}$.

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This is purely mechanical -- ask any computer algebra system. I could post Maple's answer, but the risk of transcription error is large. You are much better off doing it yourself. –  Jacques Carette Mar 15 '12 at 11:57
    
ok,thanks,I will try maple. –  Remy Mar 15 '12 at 12:38
    
I just wanted to understand the underlying procedure. –  Remy Mar 15 '12 at 13:10

1 Answer 1

up vote 1 down vote accepted

This can be computed using the 'usual' rules of calculus, along with the formula $$\frac{d}{d x} \int_{f(x)}^{g(x)} h(t,x) dt = \int_{f(x)}^{g(x)}\frac{\partial}{\partial x} h(t, x)dt + g'(x) h(g(x), x) -f'(x)h(f(x), x)$$ where $'$ denotes $\frac{d}{dx}$. If $g(x)=\infty$, you need to take a limit (over a new constant); generally this means that the second term in the formula drops out unless $$\lim_{c\to\infty} h(c,x)$$ is unbounded. Then you have to work harder.

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