1
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Wow, that must have been painful! $\endgroup$ – JPi Feb 23 '14 at 23:55
  • $\begingroup$ @JPi Yes. I went back and forth before I saw the right road. Thanks. $\endgroup$ – Felix Marin Feb 23 '14 at 23:56
  • $\begingroup$ @FelixMarin Thanks a lot. $\endgroup$ – sky-light Feb 24 '14 at 9:38
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – sky-light Feb 23 '14 at 22:09
  • 1
    $\begingroup$ 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. $\endgroup$ – JPi Feb 23 '14 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.