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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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A notational note: one uses $\gamma(a,x)$ for the "lower" version, $\Gamma(a,x)$ for the "upper" version, and different notation altogether for the "regularized" versions. Which one are you really interested in? –  J. M. Apr 29 '11 at 19:14
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In any event... formulae 8.14.3 and 8.14.4 here (Mellin transforms of incomplete gamma functions) are useful. –  J. M. Apr 29 '11 at 19:26
    
You haven't answered my question. Which of the lower and the upper ones do you want? –  J. M. Apr 29 '11 at 19:56
    
I'm using upper incomplete gamma. In my case by gamma I mean the version normalized by $\Gamma(A)$. I guess the case with $a=2$ is the answer. –  ACAC Apr 29 '11 at 19:57
    
Okay, so the integral of $x\frac{\Gamma(a,x)}{\Gamma(a)}$ then... unfortunately I have to step out, so the answer might come a bit later. –  J. M. Apr 29 '11 at 20:00

2 Answers 2

up vote 4 down vote accepted

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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The quoted result is derivable from the relationship between the incomplete gamma function and the confluent hypergeometric functions. –  J. M. Apr 30 '11 at 0:32
    
is there any condition $A$? Suppose $A=1$ and $a \neq 2$. The gamma would give you $\Gamma(1,x)=e^{-x}$ and you would get, as a final result, $\Gamma(a)$. However, with the proposed solution, you would obtain $\Gamma(a+A)/\Gamma(a)$ which is not the same –  ACAC Apr 30 '11 at 0:59
    
@Bob: you're absolutely right; I slipped up there. Hopefully things are now clearer. –  J. M. Apr 30 '11 at 1:03
    
what I was saying is this: suppose that within the integral you have $x^{a-1}$ instead of $x$, and that $A=1$. You know that $Q(1, x) = e^{-x}$ and, therefore, the result would be $\Gamma(a)$. If you use the formula above you obtain $\Gamma(a+1)/\Gamma(a)$, or am I missing something? –  ACAC Apr 30 '11 at 1:18
    
@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. –  J. M. Apr 30 '11 at 1:24

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 above –  ACAC Apr 29 '11 at 22:33

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