Other representation of Gamma Incomplete function I have a question regarding Gamma Incomplete function:
In the "Table of Integrals, Series, and Products, Seventh Edition" equation $8.353.3$ page $900$, there is a defenition for the incomplete gamma function in the case $a < 1$ and $x > 0$
$$ \Gamma(a,x)=\frac{\rho^{-x}x^{a}}{\Gamma(1-a)} \int_0^\infty \frac{e^{-t} t^{-a}}{x+t} dt$$
what is $ \rho $ in the above equation? I thought this might be a but I tried to derive the above formula but I don't got the same result.
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\Gamma\pars{a,x} = {\expo{-x}x^{a} \over \Gamma\pars{1 - a}}
\int_{0}^{\infty}{\expo{-t}t^{-a} \over x + t}\,\dd t:\ {\Large ?}}$

\begin{align}
&\color{#f00}{\Gamma\pars{a,x}\Gamma\pars{1 - a}}=
\int_{x}^{\infty}\dd t\,t^{a - 1}\expo{-t}\int_{0}^{\infty}\dd t'\,
t'^{\pars{1 - a} - 1}\expo{-t'}
\\[3mm]&=\int_{x^{1/2}}^{\infty}\dd t\,\pars{2t}t^{2a - 2}\expo{-t^{2}}
\int_{0}^{\infty}\dd t'\,\pars{2t'}t'^{-2a}\expo{-t'^{2}}
\\[3mm]&=4\int_{0}^{\infty}\int_{0}^{\infty}\Theta\pars{t - x^{1/2}}t^{2a - 1}
t'^{1 - 2a}\expo{-\pars{t^{2} + t'^{2}}}\,\dd t\,\dd t'
\\[3mm]&=4\int_{0}^{\pi/2}\dd\theta\int_{0}^{\infty}\dd r\,r\,
\Theta\pars{r\cos\pars{\theta} - x^{1/2}}r^{2a - 1}\cos^{2a - 1\pars{\theta}}
r^{1 - 2a}\sin^{1 - 2a}\pars{\theta}\expo{-r^{2}}
\\[3mm]&=4\int_{0}^{\infty}\dd r\,r\expo{-r^{2}}\int_{0}^{\pi/2}\dd\theta\,
\Theta\pars{\cos\pars{\theta} - {x^{1/2} \over r}}\cos^{2a -1}\pars{\theta}
\sin^{1 - 2a}\pars{\theta}
\\[3mm]&=2\int_{0}^{\infty}\dd t\,\expo{-t}\int_{0}^{\pi/2}\dd\theta\,
\Theta\pars{\cos\pars{\theta} - \bracks{x \over t}^{1/2}}\cos^{2a -1}\pars{\theta}
\sin^{1 - 2a}\pars{\theta}
\\[3mm]&=2\int_{0}^{\infty}\dd t\,\expo{-t}\int_{0}^{1}\dd t'\,
\Theta\pars{t' - \bracks{x \over t}^{1/2}}t'^{2a - 1}\pars{1 - t'^{2}}^{-a}
\\[3mm]&=2\int_{0}^{\infty}\dd t\,\expo{-t}\int_{0}^{1}\dd t'\,\half\,t'^{-1/2}
\Theta\pars{t' - {x \over t}}t'^{a - 1/2}\pars{1 - t'}^{-a}
\\[3mm]&=\int_{0}^{\infty}\dd t\,\expo{-t}\int_{0}^{1}\dd t'\,
\Theta\pars{tt' - x}t'^{a - 1}\pars{1 - t'}^{-a}
=\int_{0}^{1}\dd t'\,t'^{a - 1}\pars{1 - t'}^{-a}\int_{x/t'}^{\infty}\dd t\,\expo{-t}
\\[3mm]&=\int_{0}^{1}\dd t'\,t'^{a - 1}\pars{1 - t'}^{-a}\expo{-x/t'}
=\int_{\infty}^{1}t^{1 - a}\pars{1 - {1 \over t}}^{-a}\expo{-xt}\,
\pars{-\,{\dd t \over t^{2}}}
\\[3mm]&=\int_{1}^{\infty}t^{-1}\pars{t - 1}^{-a}\expo{-xt}\,\dd t
=\int_{0}^{\infty}{t^{-a} \over t + 1}\expo{-x\pars{t + 1}}\,\dd t
=\expo{-x}\int_{0}^{\infty}{\expo{-xt}t^{-a} \over t + 1}\,\dd t
\\[3mm]&=
\color{#f00}{\expo{-x}x^{a}\int_{0}^{\infty}{\expo{-t}t^{-a} \over t + x}\,\dd t}
\end{align}

Then
$$\color{#00f}{\large%
\Gamma\pars{a,x} = {\expo{-x}x^{a} \over \Gamma\pars{1 - a}}
\int_{0}^{\infty}{\expo{-t}t^{-a} \over x + t}\,\dd t}
$$
A: It should be $e$.  See the Handbook of Mathematical Functions, 8.6.4.
