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Is the expectation of the (upper/lower) incomplete gamma function known?

$\int_0^{+\infty} x \Gamma(A, x) \mathrm dx$

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    no problem, but I think its 8.14.4 with $a=2$. You gave me the reference. So if you post a quick answer I will accept it anyways.2011-04-29

2 Answers 2

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Using formula 8.14.4 in the DLMF (the Mellin transform of the upper incomplete gamma function $\Gamma(b,x)$) and specializing that formula to the case $a=2$ gives the result

$\int_0^{+\infty} x \Gamma(A, x) \mathrm dx=\frac{\Gamma(A+2)}{2}$

which is valid for $A > -2$. Alternatively, using the regularized form $Q(b,x)=\frac{\Gamma(b,x)}{\Gamma(b)}$, we have the result

$\int_0^{+\infty} x Q(A, x) \mathrm dx=\frac{A(A+1)}{2}$

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    @Bob: if you couch 8.14.4 in terms of the regularized form, the result becomes $\frac{\Gamma(a+b)}{a\Gamma(b)}$. You get the result you're saying by substituting appropriate values.2011-04-30
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It's not an expectation as stated because the supposed pdf does not integrated to unity. If $A+2>0$, the integral is $\Gamma(A+2)/2$

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    apologies for the mistake, but the integral I have to compute is the one above2011-04-29