I have an integral of the form
$$
I(a,m, k_1, ...,k_n):=\int_{0}^\infty x^m \exp(-x^2a)\Gamma(k_1, x^2)\cdot ...\cdot\Gamma(k_n, x^2)dx
$$
where $a\geq1$ and $m , k_1, ...,k_n $ are integer numbers and $\Gamma$ the upper incomplete gamma function. Is an analyticall solution possible? Maybe someone has encounterd similar integrals before. I am thankful for any advice on how to solve it.
Edit:
So far I was able to derive the formula of the integral
$$
J(a,m, k_1, ...,k_n):=\int_{0}^\infty x^m \exp(-xa)\gamma(k_1, x)\cdot ...\cdot\gamma(k_n, x)dx
$$
(I substituted $x$ for $x^2$ compared to $I(a, m , k_1, ...,k_n)$ above)
where $\gamma$ is the $\bf{lower}$ incomplete gamma function. For that, one only needs to use the power series
$$
\gamma(s, x) = x^s e^{-x}\sum_{k=0}^\infty\frac{x^k}{\Gamma(s+k+1)}
$$
The functions $\gamma$ and $\Gamma$ are for simplicity the regularized ones. With the power series the $J$ integral is only a sum of multiple integrals of the form $\int_0^{\infty} x^u\exp(-x(a+r))$ with $u$ and $r$ integers, the integral is an usual Gamma function.
The result turns out to be $n$ sums over
$$
J(a,m, k_1, ...,k_n) = \sum_{j_1, ...j_n = 0}^{\infty} \frac{\gamma(1 + m + \sum_{i = 1}^n k_i + j_i)}{\prod_{i=0}^n\gamma(k_i + j_i + 1)(a + n)^{1 + m + \sum_{i = 1}^n k_i + j_i}}
$$
This converges quite fast.
The problem is because $\Gamma(k, x) = 1 - \gamma(k, x)$ I would have to multipliy the integral $J$ $n$-times out. Because $(1 - \gamma(k_1, x))(1 - \gamma(k_2, x))(1 - \gamma(k_3, x))... = 1 - \gamma(k_1, x) + \gamma(k_1, x)\gamma(k_2, x).. $.
This Works for small $n$ but for big ones the formula is not usable ($2^{100}$ is already $ \propto 10^{30}$)
I am slowly losing hope on an usable formula.
Quite many years ago, I encounterd exactly the same problem and, alas, I never found any solution except for the simplest case $$I_1=\int_0^\infty x^m\, e^{-a x^2}\,\Gamma \left(k_1,x^2\right)\,dx$$ $$I_1=\frac{\Gamma \left(\frac{m+1+2k_1}{2}\right)}{m+1} \, _2F_1\left(\frac{m+1}{2},\frac{m+1+2k_1}{2};\frac{m+3}{2};-a\right)$$
I shall follow your question since it is an interesting problem which left me frustrated for too long.