# 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.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\Gamma\pars{a,x} = {\expo{-x}x^{a} \over \Gamma\pars{1 - a}} \int_{0}^{\infty}{\expo{-t}t^{-a} \over x + t}\,\dd t:\ {\Large ?}}$

\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 $$\color{#00f}{\large% \Gamma\pars{a,x} = {\expo{-x}x^{a} \over \Gamma\pars{1 - a}} \int_{0}^{\infty}{\expo{-t}t^{-a} \over x + t}\,\dd t}$$

• Wow, that must have been painful! – JPi Feb 23 '14 at 23:55
• @JPi Yes. I went back and forth before I saw the right road. Thanks. – Felix Marin Feb 23 '14 at 23:56
• @FelixMarin Thanks a lot. – sky-light Feb 24 '14 at 9:38

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

• Do you mean "Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 1970" By Milton Abramowitz, Irene A. Stegun? Page 334 equation 8.6.4 is it related to my case? – sky-light Feb 23 '14 at 22:09
• No, the NIST Handbook of Mathematical Functions by Olver et al.. It is identical to your case except that $\rho$ in your formula is $e$ in theirs. – JPi Feb 23 '14 at 23:48