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How can I integrate this function? It's originated by an exponential prior and a poisson likelihood.

$\int_{0}^{\infty}\lambda^{x}e^{-2\lambda}d\lambda$

Filipe Ferminiano
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1 Answers1

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This is related to the gamma function:

$$\Gamma(z) = \int_0^{\infty} dt \, t^{z-1} e^{-t}$$

In your case, sub $t=2 \lambda$, $d\lambda = t/2$ and get

$$2^{-(x+1)} \int_0^{\infty} dt \, t^x \, e^{-t} = 2^{-(x+1)} \Gamma(x+1) = \frac{x}{2^{x+1}} \Gamma(x)$$

Ron Gordon
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    why can't you write it as $\int_0^{\infty} t^{z-1} e^{-t} dt$? where did your notation come from anyway? – shimee May 27 '13 at 19:58
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    @shimee: I can, I just prefer not to. It's from my background in optics where this notation is commonplace and I used it throughout my career. I am just more comfortable with it, and I think it is more clear, especially with multiple integrals. – Ron Gordon May 27 '13 at 20:00
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    @shimee, you'll sometimes see integrals with the differential written first and the integrands coming after. Depends on the field, depends on the preferred conventions of the writer... – J. M. ain't a mathematician May 27 '13 at 20:01