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 $s>1$
Let $\lambda> 0$ and $\beta = \lambda (\alpha - 1)$. Then
$$
\Gamma (1 - \alpha ,\beta ) \sim \frac{{\beta ^{1 - \alpha } \mathrm{e}^{ - \beta } }}{{\alpha - 1}}\sum\limits_{n = 0}^\infty {\frac{{p_n (\lambda )}}{{(\lambda + 1)^{2n + 1} }}\frac{1}{{(\alpha - 1)^n }}}
$$
as $\alpha\to+\infty$ uniformly in $\lambda > 0$. Here $p_0(\lambda)=1$ and
$$
p_n (\lambda ) = \lambda (1 + \lambda )p'_{n - 1} (\lambda ) - (2n - 1)\lambda p_{n - 1} (\lambda )
$$
for $n\geq 1$. These are polynomials in $\lambda$ of degree $n$.
To achive the best numerical accuracy, stop the series after about $\left\lfloor {\alpha \sqrt {(\lambda + \log \lambda + 1)^2 + \pi ^2 } } \right\rfloor$ terms. See this paper for more details.
Best Answer
For $\operatorname{Re}(\alpha) < 1$ and $\beta > 0$ the function $$ f \colon \mathbb{R}^+ \to \mathbb{C}\, , \, f(x) = \int \limits_\beta^\infty \mathrm{e}^{- x \theta} \frac{1}{\theta} \left(\frac{\theta}{\beta}-1\right)^{-\alpha} \, \mathrm{d} \theta \, , $$ is well-defined and differentiable. Using the change of variables $\theta = \beta \left(1 + \frac{\phi}{\beta x}\right)$ we find for $x > 0$ $$ f'(x) = - \int \limits_\beta^\infty \mathrm{e}^{- x \theta} \left(\frac{\theta}{\beta}-1\right)^{-\alpha} \, \mathrm{d} \theta = - \beta^{\alpha} x^{\alpha-1} \mathrm{e}^{-\beta x} \int \limits_0^\infty \phi^{-\alpha} \mathrm{e}^{- \phi} \, \mathrm{d} \phi = - \beta^{\alpha} x^{\alpha-1} \mathrm{e}^{-\beta x} \, \Gamma(1-\alpha) \, .$$ Since $\lim_{x \to \infty} f(x) = 0$ , we obtain \begin{align} f(x) &= - \int \limits_x^\infty f'(y) \, \mathrm{d} y = \Gamma(1-\alpha) \beta^{\alpha} \int \limits_x^\infty y^{\alpha-1} \mathrm{e}^{- \beta y} \, \mathrm{d} y = \Gamma(1-\alpha) \int \limits_{\beta x}^\infty z^{\alpha-1} \mathrm{e}^{- z} \, \mathrm{d} z \\ &= \Gamma(1-\alpha) \Gamma(\alpha, \beta x) \, \end{align} for $x > 0$ .
The Laplace transform of the gamma function does not exist, since $\Gamma$ grows faster than any exponential function.