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I'm interested (ref) in the following integral

$$I(m,d)=\int_0^{\infty} \left( \frac{\Gamma(m,x)}{\Gamma(m)} \right)^d dx=\frac{1}{((m-1)!)^d}\int_0^{\infty} \Gamma(m,x)^d dx$$

where $\Gamma(m,x)$ is the (upper) incomplete gamma function, $m,d$ are positive integers.

In particular, I'm interested in $d=3$.

Exact solutions, approximations or asymptotics (for $m \to \infty$) are appreciated.

Numerically, it seems that $I(m,3) = m - a \sqrt{m} +O(1)$ with $a \approx 0.835$

Some values for $d=3$

2   0.96296
3   1.68313
4   2.44942
5   3.24473
10  7.44823
20  16.3304
50  44.1225
100 91.6395
200 188.1311
300 285.4399
400 383.1715
500 481.1731

In case this helps: Asymptotic expansions for the incomplete gamma function...

leonbloy
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    Maybe this helps:$$\int_0^{\infty } \Gamma (m,x)^3 , dx=\Gamma (m)^3 \sum _{s=0}^{m-1} \sum _{j=0}^{m-1} \sum _{k=0}^{m-1} \frac{3^{-1-j-k-s} \Gamma (1+j+k+s)}{j! k! s!}$$ – Mariusz Iwaniuk Sep 26 '19 at 16:19
  • @MariuszIwaniuk I'm not sure it helps, but it looks like an interesting alternate way to rediscover my answer to the linked question. Does it have a simple justification? – leonbloy Sep 26 '19 at 16:40
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    Put this: http://functions.wolfram.com/GammaBetaErf/Gamma2/06/01/04/01/02/0004/ sum to integral. – Mariusz Iwaniuk Sep 26 '19 at 16:46
  • If I may ask : up to which value of $m$ did you perform the numerical integration ? I face incredible problems with this integral. Cheers. – Claude Leibovici Sep 30 '19 at 12:45
  • @ClaudeLeibovici In Maxima the integral can be evaluated for $m $ around 50, IIRC . For larger values, I used my recursive code in the linked answer (as it is, it works for $m\le 127$, but it could easily be extended for larger values). – leonbloy Sep 30 '19 at 15:02
  • @leonbloy. To me, it looks more linear than what you give. If I am not mistaken, for $m=100$, the result is $91.639$. Am I corect ? – Claude Leibovici Sep 30 '19 at 15:15
  • Correct. I added some values – leonbloy Sep 30 '19 at 18:10
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    Indeed $\lim\limits_{m\to\infty}\frac{I(m,d)}{m}=1$ for any $d$, and $a=\frac{3}{2\sqrt\pi}$ (for $d=3$). I'm still trying to make my arguments clear, and to find $a=a(d)$ for $d\neq 3$. – metamorphy Mar 22 '20 at 16:19
  • Do an inverse substitution and use the $(50)$ to $(52)$ series expansion, assuming it converges for the entire integral, for an exact answer. Would this answer be wanted? – Тyma Gaidash Sep 09 '23 at 12:25

1 Answers1

-1

Answered in MathOverflow.

The asymptotic expansion for general $d$ is shown to be

$$I(m,d) \sim m - m^{-1/2} a_d$$

where $a_d$ has a rather complex form (see the linked answer).

For $d=3$, $a_3=\frac{3}{2\sqrt{\pi}}=0.846284\cdots$ (in agreement with metamorphy's comment).

leonbloy
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