Laplace Transform of Incomplete Gamma Function

While looking through a table of integral transforms ($$178$$,$$(30)$$) I have come across the Laplace Transform of the Incomplete Gamma Function which is given by

$$\mathcal{L}\{\Gamma(\nu,at\}(p)~=~\Gamma(\nu)p^{-1}\left(1-\left(1+\frac{p}{a}\right)^{-\nu}\right)~~~~~~\text{Re}~\nu>-1,\,\text{Re}~p>-\text{Re}~a$$

I was a little bit confused that it seems like there is no transform at all for the Gamma Function itself. This was the original reason why I have looked through this table. I am not that familiar with checking the convergence of an integral I do not understand why such an transform does not exist.

Therefore, my first question is how one could show that the Gamma Function does not posses a Laplace Transform - and to go further whether it is true or not that there is no integral kernel at all which provides a transform for the Gamma Function.

I guess this has something to do with the fact that the Gamma Function is not bounded, as far as I know, but I would leave it to someone else to form a proof.

Back to equation from above. I tried to derive this formula by myself and failed. First, I decided to set $$a=1$$ to understand the process at all and do not make it even more complicated by having an extra factor in there. To be precise this was my attempt so far

\begin{align} \mathcal{L}\{\Gamma(\nu,t\}(p)~&=~\int_0^{\infty}~\Gamma(\nu,t)~e^{-st}~\mathrm{d}t\\ &=~\int_0^{\infty}~\left[\int_{\nu}^{\infty}x^{t-1}~e^{-x}~\mathrm{d}x\right]~e^{-st}~\mathrm{d}t\\ &=~\int_{\nu}^{\infty}\int_0^{\infty}~x^{t-1}~e^{-st}~\mathrm{d}t~e^{-x}~\mathrm{d}x\\ &=~\int_{\nu}^{\infty}\left[\frac{e^{t(\log(x)-z)}}{x(\log(x)-z)}\right]_0^{\infty}~e^{-x}~\mathrm{d}x \end{align}

I am not not even sure if I am allowed to use Fubini's-Tonelli's-Theorem here nor how to calculate the definite integral. Since $$\frac1{x(\log(x)-z)}$$ can be considered as constant in this context I am left with the limit

$$\lim_{t\to\infty}\frac{x^t}{e^{zt}}$$

And I have no clue how to approach to this. Both functions are exponetials and so l'Hospital does not help at all. I do not know how to evaluate a limit like this. Therefore this is my second question. Even by assuming that the limit is $$0$$ I am left with the definite integral

$$\int_{\nu}^{\infty}\frac1{x(z-\log(x))}~e^{-x}~\mathrm{d}x$$

which I cannot evaluate either. Therefore I guess I made a crucial mistake somewhere or this is the wrong attempt to this problem.

Finally I would like to ask for both, a derivation and a proof, for the Laplace Transform of the Incomplete Gamma Function given by the first formula. Furthermore I would be pleased if someone could point out where I did made a mistake so that my approach did not worked out.

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:

1. $\alpha>s>1$
2. $\alpha\in\mathbb Z^-\cup\{0\}$ and $s>1$
• Thank you for the detailed answer. I should have done more research on the Incomplete Gamma Function to prevent such a mistake concerning the defintion. Since I guess one of the main reasons why my attempt failed was the wrong defintion. Actually there is a transform, the so called Weierstrass Transform, with a Gaussian as Kernel. To be exact $$K(x,y)~=~\large\frac{e^{-\frac{(x-y)^2}{2}}}{\sqrt{4\pi}}$$. Therefore it could be that there exist maybe even a Weierstrass Transform of the Gamma Function? Aug 10, 2018 at 7:09
• I just realized I have made a mistake. I am forced to correct it in another comment since my edit time ran out. The denominator of the fraction within the exponential should be a $4$ instead of a $2$. Furthermore I guess my defintion was only wrong concerning the fact that I have swapped the role of the first and second argument for no obvious reason since the right one was used in the first answer to this post for example. Aug 10, 2018 at 7:27

The identity follows from $$\Gamma(\nu, a t) = \Gamma(\nu) - a^\nu \int_0^t \tau^{\nu - 1} e^{-a \tau} d\tau, \\ \operatorname{Re} \nu > 0 \land t > 0$$ and $\mathcal L[\int_0^t f(\tau) d\tau] = p^{-1} \mathcal L[f]$: $$\mathcal L[\Gamma(\nu, a t)] = \mathcal L[\Gamma(\nu)] - \frac {a^\nu \mathcal L[t^{\nu - 1} e^{-a t}]} p = \\ \frac {\Gamma(\nu) (1 - a^\nu (p + a)^{-\nu})} p.$$ Note that the formula with $(1 + p/a)^{-\nu}$ is valid only for $\operatorname{Re} a > 0 \lor \nu \in \mathbb N$.

Then the result can be extended to $\operatorname{Re} \nu > -1$ by analytic continuation.