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I tried a few ways (integral by parts, expanding), but I'm unable to compute this integral.

$$\int_0^\infty\frac{\lambda^{n}e^{-\lambda}}{(\lambda + b)^2}\, \text{d}\lambda$$

$n \geq 0, b \geq 0$

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  • $\begingroup$ Whether it is helpful or not, I see the $\lambda^ne^{-\lambda}$ and think Gamma function. $\endgroup$ – Ethan Hunt Apr 19 '16 at 14:56
  • $\begingroup$ What are $b$, $n$? $\endgroup$ – Yuriy S Apr 19 '16 at 15:15
  • $\begingroup$ Basically I am trying to find the marginal likelihood of poisson (rate parameter) with the prior $(\lambda +b)^{-2}$. n is the number of events observed. $\endgroup$ – akashrajkn Apr 19 '16 at 15:46
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If $n$ is a non-negative integer you can put $\mu = \lambda + b $ so that the integral becomes $$ e^b \int_b^{\infty} \frac{(\mu-b)^n e^{-\mu}}{\mu^2} d\mu = e^b \sum_{k=0}^n \binom{n}{k}(-b)^{n-k} \int_b^{\infty} \mu^{k-2} e^{-\mu} d\mu$$ Then, except for when $k=0$ or $1$, the integrals in the sum can be solved using integration by parts whereas $$ \int_b^{\infty} \frac{e^{-\mu}}{\mu}d\mu = -\text{Ei}(-b) \,\,\,,\,\,\, \int_b^{\infty} \frac{e^{-\mu}}{\mu^2}d\mu = \text{Ei}(-b) + \frac{e^{-b}}{b}$$

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    $\begingroup$ The integrals in the sum represent the Incomplete Gamma functon, namely $$\int_b^{\infty} \mu^{k-2} e^{-\mu} ~d \mu=\Gamma(k-1,b)$$ $\endgroup$ – Yuriy S May 5 '16 at 14:21
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Even for $n=0$, the integral requires special functions, see Wolfram Alpha, so there probably is no simple way of computing it.

For $n=0$, the result is $$-e^b \text{Ei}(-b-\lambda )-\frac{e^{-\lambda }}{b+\lambda }$$

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