I am trying to calculate the following integral:

$$ \int_0^{\infty}e^{-\beta x}\gamma(\alpha,\theta x)dx $$ where all parameters are positive.

Any help , Thanks!

  • $\begingroup$ Integrate by parts, recalling the Leibniz rule for differentiating under the integral to calculate $\gamma '$. $\endgroup$ – stochasticboy321 Jan 29 '16 at 22:54
  • $\begingroup$ Since $\beta$ is positive, you might at least perform a Taylor expansion of the exponential, in order to get $$\int x^k\ \gamma(\alpha, \theta x)\ \text{d}x$$ (In which I left out the sum of the expansion). Then you may try the by parts integration. $\endgroup$ – Turing Jan 29 '16 at 22:56
  • $\begingroup$ @KimPeek isn't that needlessly complicating things? For starters, $e^{-\beta x}$ is easy to integrate over $[0,\infty)$, and further, it isn't obvious to me that the integral you have written is finite over the domain for any given $k$. $\endgroup$ – stochasticboy321 Jan 29 '16 at 23:14
  • $\begingroup$ @stochasticboy321 You may be right, but a starter wouldn't treat the incomplete gamma function lol. But sure, the direct by part integration might work better.. As usual I find the most strange paths :D $\endgroup$ – Turing Jan 29 '16 at 23:19
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    $\begingroup$ @KimPeek Oh dear, there might have been a 'cross the pond issue - "for starters" merely means "to begin with". As to the mathematics, the issue is that one would have to retain the integrals in indefinite form, and sum the series before applying the limits. This would involve ghastly bookkeeping that goes away in the direct integration of the exponential. The method's certainly valid, and useful in many cases, but not in this particular one. $\endgroup$ – stochasticboy321 Jan 30 '16 at 1:51

The integral definition of $\gamma(\alpha,X)$ leads to a double integral which can be expressed on closed form :

enter image description here

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