# Integrating the lower incomplete gamma $\int_0^\infty x^{a-1}e^{-s x} \gamma(b,x) \mathrm{d}x$

I need to prove that

$$\int_0^\infty x^{a-1}e^{-s x} \gamma(b,x) \mathrm{d}x = \frac{\Gamma(a+b)}{a(1+s)^{a+b}}F(1,a+b,1+b, 1/(1+s))$$ where $F(a,b,c;x)$ is the hypergeometric function. To show this, I tried the following:

\begin{eqnarray} \int_0^\infty x^{a-1}e^{-s x} \gamma(b,x) \mathrm{d}x &=& \int_0^\infty \int_0^x x^{a-1}e^{-s x} y^{b-1}e^{-y} \mathrm{d}y\mathrm{d}x \end{eqnarray} Substitute $y = x \cdot z$ such that $\mathrm{d}y = x \cdot \mathrm{d}z$, we have: \begin{eqnarray} \int_0^\infty \int_0^1 x^{a+b-1}e^{-x(s+z)} z^{b-1}\mathrm{d}z\mathrm{d}x &=& \int_0^\infty \int_0^1 (x(s+z))^{a+b-1}e^{-x(s+z)} \frac{z^{b-1}}{(s+z)^{a+b-1}}\mathrm{d}z\mathrm{d}x \end{eqnarray} Setting $\frac{u}{(s+z)} = x$ and $\mathrm{d}x = \frac{1}{(s+z)}\mathrm{d}u$, and changing the integration order, we obtain

\begin{eqnarray} \int_0^\infty \int_0^1 (x(s+z))^{a+b-1}e^{-x(s+z)} \frac{z^{b-1}}{(s+z)^{a+b-1}}\mathrm{d}z\mathrm{d}x&=& \int_0^1 \int_0^\infty u^{a+b-1}e^u \frac{z^{b-1}}{(s+z)^{a+b}}\mathrm{d}u\mathrm{d}z \\ &=& \Gamma(a+b) \int_0^1 \frac{z^{b-1}}{(s+z)^{a+b}}\mathrm{d}z \end{eqnarray}

I know that there is an integral representation to the hypergeometric function $$F(a,b,c;x) = \frac{1}{\Gamma(b)\Gamma(c-b)}\int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-x t)^a}$$ however, I couldn't reach further and obtain the required form of the hypergeometric function.

What am I missing? Any ideas how to solve this? suggestions and clues are much appreciated.

Once you correctly get $$I=\Gamma(a+b)\int_{0}^{1}\frac{z^{b-1}}{(s+z)^{a+b}}\,dz$$ just replace $z$ with $1-z$ to get: $$I = \Gamma(a+b) \int_{0}^{1}\frac{(1-z)^{b-1}}{(s+1-z)^{a+b}} = \frac{\Gamma(a+b)}{(s+1)^{a+b}}\int_{0}^{1}\frac{(1-z)^{b-1}}{\left(1-\frac{1}{s+1}\,z\right)^{a+b}}\,dz$$ and recognize the integral representation of a $\phantom{}_2 F_1$.
• Thanks once again @Jack ! Is there an identity of the hypergeometric function used to obtain the theorem ? In this integral representation we have $F(a+b,1,b+1, 1/(s+1)$. How can we get to $F(1,a+b,1+b, 1/(1+s))/a$. ?
• @sefi: $$\phantom{}_{2} F_1(a,b;c;z)=\phantom{}_{2} F_1(b,a;c;z).$$ Jun 14 '16 at 6:20
• Great, thanks @jack. I thought that there's an identity here since the $1/a$ in the theorem. So, from where do we get this $1/a$ in the solution?
• @sefi: that $a$ comes from the $\Gamma$ factors in the integral representation of the $\phantom{}_2 F_1$. Jun 14 '16 at 6:28