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DLMF 5.9.1 reproduces the result from exercise 1.1 from ch 2.1 of Olver's Asymptotic and Special Functions:

$$\frac{1}{\mu}\Gamma\left(\frac{\nu}{\mu}\right)\frac{1}{z^{\nu/\mu}}=\int_{0}^% {\infty}\exp\left(-zt^{\mu}\right)t^{\nu-1}\mathrm{d}t$$

That result is valid when $\Re\nu, \mu, \Re z>0$.

Is there a simple generalization when $\nu, \mu<0,\ z>0,\ \nu,\mu,z\in\mathbb{R}$? It seems to me that in that case, the formula becomes $-\frac{1}{\mu}\Gamma\left(\frac{\nu}{\mu}\right)\frac{1}{z^{\nu/\mu}}$, but is there a simple way to show it?

The motivation for this came from statistics, where one had to show that

$$\int_{0}^{\infty}\frac{1}{\Gamma(a)\ b^{a}}\ \exp\left(-\frac{1}{b}t^{-1}\right) t^{-a-1}\mathrm{d}y=1 \text{ (i.e. is a PDF)},$$ with $a,b>0$.

Applying DLMF 5.9.1 while ignoring the constraints yields -1 instead of +1.

Kuba hasn't forgotten Monica
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1 Answers1

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Note that \begin{align*} \int_0^{ + \infty } {\exp \left( { - zt^{ - \alpha } } \right)t^{ - \beta - 1} dt} & \mathop = \limits^{t = 1/s} - \int_{ + \infty }^0 {\exp \left( { - zs^\alpha } \right)s^{\beta - 1} ds} \\ & \;\,= \int_0^{ + \infty } {\exp \left( { - zs^\alpha } \right)s^{\beta - 1} ds} = \frac{1}{\alpha }\Gamma \left( {\frac{\beta }{\alpha }} \right)\frac{1}{{z^{\beta /\alpha } }} \end{align*} when $\alpha$, $\beta>0$. Take $\alpha=-\mu$ and $\beta=-\nu$, with $\mu$, $\nu<0$ and you have your result.

Gary
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