5
$\begingroup$

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...

$\endgroup$
9
  • 1
    $\begingroup$ Maybe this helps:$$\int_0^{\infty } \Gamma (m,x)^3 \, dx=\Gamma (m)^3 \sum _{s=0}^{m-1} \sum _{j=0}^{m-1} \sum _{k=0}^{m-1} \frac{3^{-1-j-k-s} \Gamma (1+j+k+s)}{j! k! s!}$$ $\endgroup$ Sep 26, 2019 at 16:19
  • $\begingroup$ @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? $\endgroup$
    – leonbloy
    Sep 26, 2019 at 16:40
  • 1
    $\begingroup$ Put this: functions.wolfram.com/GammaBetaErf/Gamma2/06/01/04/01/02/0004 sum to integral. $\endgroup$ Sep 26, 2019 at 16:46
  • $\begingroup$ If I may ask : up to which value of $m$ did you perform the numerical integration ? I face incredible problems with this integral. Cheers. $\endgroup$ Sep 30, 2019 at 12:45
  • 1
    $\begingroup$ 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 $d\neq 3$. $\endgroup$
    – metamorphy
    Mar 22, 2020 at 16:19

1 Answer 1

-1
$\begingroup$

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).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .