Your definition of incomplete Gamma function is wrong.
As an alternative definition of incomplete Gamma function:
$$\Gamma(v,at)=\int^\infty_1(at)^vu^{v-1}e^{-atu}du=a^vt^v\int^\infty_1u^{v-1}e^{-atu}du$$
Your original integral $I$ becomes
$$I=\int^\infty_0a^vt^v e^{-pt}\int^\infty_1u^{v-1}e^{-atu}du dt$$
By Fubini's theorem (the integrand is always positive),
$$I=a^v\int^\infty_1 u^{v-1} \underbrace{\int^\infty_0 t^ve^{-pt}e^{-atu}dt}_{=I_1} du$$
$$I_1=\int^\infty_0t^ve^{-(p+au)t}dt$$
Let $g=(p+au)t$,
$$I_1=\int^\infty_0\frac{g^v}{(p+au)^{v+1}}e^{-g}dg=\frac{\Gamma(v+1)}{(p+au)^{v+1}}$$
$$I=a^v\int^\infty_1 u^{v-1}\frac{\Gamma(v+1)}{(p+au)^{v+1}}du=a^v\Gamma(v+1)\int^\infty_1\left(\frac{u}{p+au}\right)^{v+1}\frac1{u^2}du$$
Substitute $h=\frac1u$,
$$I=a^v\Gamma(v+1)\int^1_0\frac1{(ph+a)^{v+1}}dh=a^v\Gamma(v+1)\cdot\frac{-1}{pv}\left((p+a)^{-v}-a^{-v}\right)$$
By recognizing $\Gamma(v+1)=v\Gamma(v)$ and further simplifying,
$$I=\frac{\Gamma(v)}p\left(1-\left(1+\frac pa\right)^{-v}\right)$$
An integral transform of Gamma function rarely exist, because Gamma function grows too rapidly. It grows even faster than exponential growth, that's why it does not have a Laplace transform. I'm not sure if a kernel like $e^{-x^2}$ would be okay.
Note that when we try to do a Laplace transform of Gamma function (incomplete or complete) with respect to the first argument, we always fail. If we do it w.r.t. to the second argument (for incomplete Gamma function), we get something useful, which is listed in your table.
Due to $$\lim_{x\to\infty}\frac{\Gamma(x)x^\alpha}{\Gamma(x+\alpha)}=1$$, I can 'invent' a kernel such that the transform for Gamma function exists.
$$\mathcal{T}_\alpha\{f\}(s)=\int^\infty_0 f(t)\cdot\frac{t^{\alpha-s}}{\Gamma(t+\alpha)}dt$$
$\mathcal{T}_\alpha\{\Gamma\}(s)$ exists if one of the following condition is satisfied:
- $\alpha>s>1$
- $\alpha\in\mathbb Z^-\cup\{0\}$ and $s>1$
$Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].
$
I'll blindly try
to reverse the order of summation
and see what happens.
$\begin{array}\\
S(u, v)
&=\sum_{k=0}^\infty \sum_{l=0}^k \frac{u^lv^k}{k!l!}\\
&=\sum_{l=0}^\infty\sum_{k=l}^\infty \frac{u^lv^k}{k!l!}\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=l}^\infty \frac{v^k}{k!}\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}(e^v-\sum_{k=0}^{l-1} \frac{v^k}{k!})\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}e^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1} \frac{v^k}{k!}\\
&=e^ue^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1} \frac{v^k}{k!}\\
&=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}(\sum_{k=0}^{l} \frac{v^k}{k!}-\frac{v^l}{l!})\\
&=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l} \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{u^l}{l!}\frac{v^l}{l!}\\
&=e^{u+v}-\sum_{l=0}^\infty\sum_{k=0}^{l}\frac{u^l}{l!} \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{(uv)^l}{l!^2}\\
&=e^{u+v}-S(v, u)+I_0(2\sqrt{uv})
\\
\end{array}
$
where
$I_0$
is the modified Bessel function
of the first kind.
So this isn't a evaluation
but we get the relation
$S(u, v)+S(v, u)
=e^{u+v}+I_0(2\sqrt{uv})
$.
Then
$\begin{array}\\
Q(t)
&= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big]\\
&= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}S(at, b/a)\Big]\\
&= \frac{1}{a}\Big[e^{at+b/a}-S(at, b/a)\Big]\\
&= \frac{1}{a}\Big[e^{at+b/a}-(e^{at+b/a}-S(b/a, at)+I_0(2\sqrt{(at)(b/a)}))\Big]\\
&= \frac{1}{a}\Big[S(b/a, at)-I_0(2\sqrt{tb})\Big]\\
\end{array}
$
Again,
not an evaluation,
but a possibly useful
alternative expression.
This reminds me
very much
of some work I did
over forty years ago
on the Marcum Q-function.
You might look that up
and follow the references.
You can start here:
https://en.wikipedia.org/wiki/Marcum_Q-function
Best Answer
We can use some fairly brutal estimates. On the one hand, \begin{align} \int_0^{2n} x^n e^{-x}\,dx &> \int_n^{2n} x^n e^{-x}\,dx \\ &> n^n \int_n^{2n} e^{-x}\,dx \\ &= n^ne^{-n}(1-e^{-n}) \\ & > \frac{n^n}{2e^n} \end{align} for $n \geqslant 1$.
On the other hand, with $g(x) = x^n e^{-x/2}$ we have $$g'(x) = \biggl(\frac{n}{x} - \frac{1}{2}\biggr)g(x) \leqslant 0$$ for $x \geqslant 2n$, whence \begin{align} \int_{2n}^{\infty} x^ne^{-x}\,dx &= \int_{2n}^{\infty} g(x) e^{-x/2}\,dx \\ &\leqslant g(2n) \int_{2n}^{\infty} e^{-x/2}\,dx \\ &= 2g(2n)e^{-n} \\ &= 4\cdot \frac{n^n}{2e^n}\cdot \biggl(\frac{2}{e}\biggr)^n\,. \end{align} Hence $$\frac{\Gamma(n+1,2n)}{\gamma(n+1,2n)} \leqslant 4\biggl(\frac{2}{e}\biggr)^n\,,$$ i.e. we have exponential decay.