Htykuut

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.

$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{Gammapars{a,x} = {expo{-x}x^{a} over Gammapars{1 - a}}
int_{0}^{infty}{expo{-t}t^{-a} over x + t},dd t: {Large ?}}$

begin{align}
&color{#f00}{Gammapars{a,x}Gammapars{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]&=4int_{0}^{infty}int_{0}^{infty}Thetapars{t - x^{1/2}}t^{2a - 1}
t'^{1 - 2a}expo{-pars{t^{2} + t'^{2}}},dd t,dd t'
\[3mm]&=4int_{0}^{pi/2}ddthetaint_{0}^{infty}dd r,r,
Thetapars{rcospars{theta} - x^{1/2}}r^{2a - 1}cos^{2a - 1pars{theta}}
r^{1 - 2a}sin^{1 - 2a}pars{theta}expo{-r^{2}}
\[3mm]&=4int_{0}^{infty}dd r,rexpo{-r^{2}}int_{0}^{pi/2}ddtheta,
Thetapars{cospars{theta} - {x^{1/2} over r}}cos^{2a -1}pars{theta}
sin^{1 - 2a}pars{theta}
\[3mm]&=2int_{0}^{infty}dd t,expo{-t}int_{0}^{pi/2}ddtheta,
Thetapars{cospars{theta} - bracks{x over t}^{1/2}}cos^{2a -1}pars{theta}
sin^{1 - 2a}pars{theta}
\[3mm]&=2int_{0}^{infty}dd t,expo{-t}int_{0}^{1}dd t',
Thetapars{t' - bracks{x over t}^{1/2}}t'^{2a - 1}pars{1 - t'^{2}}^{-a}
\[3mm]&=2int_{0}^{infty}dd t,expo{-t}int_{0}^{1}dd t',half,t'^{-1/2}
Thetapars{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',
Thetapars{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{-xpars{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
$$color{#00f}{large%
Gammapars{a,x} = {expo{-x}x^{a} over Gammapars{1 - a}}
int_{0}^{infty}{expo{-t}t^{-a} over x + t},dd t}
$$

It should be $e$. See the Handbook of Mathematical Functions, 8.6.4.