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This definite integral relates to Bayesian statistics and is a problem that needs to be solved in order to show that the $\text{Gamma}(\alpha,\beta)$ distribution is a (prior) conjugate family for the $\text{Poisson}(\lambda)$ distribution. Now, I know there are ways to prove this by means of proportionality arguments, and then identify the posterior distribution to make the necessary conclusion, but I want to do it the hard (but more intuitive) way. So I have,

$$ \int_0^\infty \lambda^{k+\alpha-1}\text{e}^{-(\beta+1) \lambda} \, d\lambda $$

where $k$ is an integer $\geq0$ and $\alpha, \beta > 0$. In a "normal" situation where $\lambda$ is raised to the power of an integer, I would be able to solve it by integration of parts, but know I am not so sure. Any clues?

Edit

I use the hints user NHL provided:

\begin{align} \int_0^\infty \lambda^{k+\alpha-1}\text{e}^{-(\beta+1)\lambda} \, d\lambda &= \big\{\mu = (\beta+1)\lambda,\, d\mu = (\beta+1) \, d\lambda \big\} \\[3mm]&= \frac{1}{(\beta + 1)^{k+\alpha}} \int_0^\infty \mu^{k+\alpha-1} \text{e}^{-\mu} \, d\mu \\[3mm] &= \big\{ \Gamma(z)\triangleq \int_{0}^{\infty}x^{z-1}\text{e}^{-x} \, dx\big\}\\[3mm] &= \frac{\Gamma(k+\alpha)}{(\beta + 1)^{k+\alpha}}. \end{align}

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    $\begingroup$ even if $\lambda$ is raised to a non-integer power, the integration by part works the same $\endgroup$
    – NHL
    Feb 5, 2021 at 10:20
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    $\begingroup$ $ \int_{0}^{\infty}\lambda^{0.5}\text{e}^{-\lambda}d\lambda= [-\lambda^{0.5}e^{-\lambda}]_0^\infty+0.5\int_0^\infty \lambda^{-0.5}e^{-\lambda}$. so you get $\Gamma(0.5)=0.5\Gamma(-0.5)$ $\endgroup$
    – NHL
    Feb 5, 2021 at 10:30
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    $\begingroup$ Is that the definition of the gamma function? I see! How about this case: $$\int_{0}^{\infty}\lambda^{0.5}\text{e}^{-2\lambda}$$ where we cannot use the definition directly. Is it still possible? @NHL $\endgroup$
    – SimpleProgrammer
    Feb 5, 2021 at 10:37
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    $\begingroup$ You can use a change of variable $\mu=2\lambda $ to get to the definition of the gamma function $\endgroup$
    – NHL
    Feb 5, 2021 at 10:42
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    $\begingroup$ Just a friendly reminder: if you have an answer (even one of your own) then post it as such and check it. $\endgroup$
    – user190080
    Feb 5, 2021 at 13:11

1 Answer 1

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I use the hints user NHL provided:

\begin{align} \int_{0}^{\infty}\lambda^{k+\alpha-1}\text{e}^{-(\beta+1)\lambda}d\lambda &= \big\{\mu = (\beta+1)\lambda,\; d\mu = (\beta+1)d\lambda\big\} \\[3mm]&= \frac{1}{(\beta + 1)^{k+\alpha}}\int_{0}^{\infty}\mu^{k+\alpha-1}\text{e}^{-\mu}d\mu \\[3mm] &= \big\{ \Gamma(z)\triangleq \int_{0}^{\infty}x^{z-1}\text{e}^{-x}dx\big\}\\[3mm] &= \frac{\Gamma(k+\alpha)}{(\beta + 1)^{k+\alpha}}. \end{align}

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