# Integral involving gamma incomplete function.

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...

• Maybe this helps:$$\int_0^{\infty } \Gamma (m,x)^3 \, dx=\Gamma (m)^3 \sum _{s=0}^{m-1} \sum _{j=0}^{m-1} \sum _{k=0}^{m-1} \frac{3^{-1-j-k-s} \Gamma (1+j+k+s)}{j! k! s!}$$ Commented Sep 26, 2019 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? Commented Sep 26, 2019 at 16:40
• Put this: functions.wolfram.com/GammaBetaErf/Gamma2/06/01/04/01/02/0004 sum to integral. Commented Sep 26, 2019 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. Commented Sep 30, 2019 at 12:45
• 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 $d\neq 3$. Commented Mar 22, 2020 at 16:19

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).