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…
Best Answer
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).