# Integral with incomplete gamma function

I am trying to solve this integral:

\begin{equation} \frac{1}{c^{b}}\int_{0}^{\infty} x^{n}\, e^{-a x}\, \gamma(b,c(-d+x)) \ \mathrm{d}x \end{equation}

where, $n>0$ is an integer, and $a$, $b$, $c$, $-d$ are all positive. I tried using $u = -d+x$ and then binomial expansion for the resulting $(u+d)^{n}$ term to get the following form:

\begin{equation} \frac{1}{c^{b}}\sum_{j=0}^{n} \ (\mathrm{constant}) \int_{-d}^{\infty} u^{j}\, e^{-a u}\, \gamma(b,cu) \ \mathrm{d}u \end{equation}

Which gets me to the same form of the integral in Gradshteyn and Ryzhik's Table (see entry 6.455):

\begin{equation} \int_{0}^{\infty} x^{\mu-1}\, e^{-\beta x}\, \gamma(\nu,\, \alpha x) \ \mathrm{d}x = \frac{\alpha^{\nu}\, \Gamma(\mu+\nu)}{\nu(\alpha+\beta)^{\mu+\nu}}\, {_{2}}F_{1}\left(1,\, \mu+\nu;\, \nu+1;\frac{\alpha}{\alpha+\beta}\right) \end{equation}

Which of course has different limits of integration. Any ideas on how I might solve this? I am looking for suggestions on how to approach the problem. Any methods of integration and/or helpful identities would also be great.

## 1 Answer

Solution:

\begin{equation} \frac{1}{a_{2}^{b}}\int_{0}^{\infty} x^{n}\, e^{-a_{1}x} \, \gamma(b, a_{2}(-p+x)) \ \mathrm{d}x \end{equation}

Use this fact: $\frac{\partial}{\partial x}\, \gamma(a,f(x)) = e^{-f(x)}\, f(x)^{a-1} \, f'(x) \ \mathrm{d}x$ and do integration by parts.

\begin{equation*} \begin{aligned} u &= \gamma(b, a_{2}(-p+x)) &&v = -\frac{\Gamma(n+1,a_{1}x)}{a_{1}^{n+1}}\\ du &= a_{2}^{b}\, e^{-a_{2}(-p+x)}\, (-p+x)^{b-1} \ \mathrm{d}x &&dv = x^{n}\, e^{-a_{1}x} \ \mathrm{d}x \end{aligned} \end{equation*}

Which gives:

\begin{equation} -\frac{\gamma(b, a_{2}(-p+x))\,\Gamma(n+1,a_{1}x)}{a_{1}^{n+1}a_{2}^{b}}\Bigg|_{0}^{\infty} %%%%%%%%%%%%%%%%%%%% +\frac{e^{a_{2}p}}{a_{1}^{n+1}}\int_{0}^{\infty} (-p+x)^{b-1}\, e^{-a_{2}x}\, \Gamma(n+1,a_{1}x) \ \mathrm{d}x \end{equation}

For the first term we need these limits: $\lim_{x\to\infty}\Gamma(a,b x)=0$, and $\lim_{x\to 0^{+}}\Gamma(a,b x)=\Gamma(a)$. Also, use binomial theorem to expand the $(-p+x)^{b-1}$ inside the integral.

\begin{equation} \frac{\gamma(b, -a_{2}p)\,\Gamma(n+1)}{a_{1}^{n+1}a_{2}^{b}} +\frac{e^{a_{2}p}}{a_{1}^{n+1}} \sum_{j=0}^{b-1} \binom{b-1}{j} (-p)^{b-1-j} \int_{0}^{\infty} x^{j}\, e^{-a_{2}x}\, \Gamma(n+1,a_{1}x) \ \mathrm{d}x \end{equation}

Now we can see the integral is of the form in the question:

\begin{equation*} \int_{0}^{\infty} x^{j}\, e^{-a_{2} x}\, \Gamma(n+1,\, a_{1} x) \ \mathrm{d}x = \frac{a_{1}^{n+1}\, \Gamma(j+n+2)}{(j+1)(a_{1}+a_{2})^{j+n+2}}\, {_{2}}F_{1}\left(1,\, j+n+2;\, j+2;\frac{a_{2}}{a_{1}+a_{2}}\right) \end{equation*}

So the final answer is:

\begin{equation} \frac{\gamma(b, -a_{2}p)\,\Gamma(n+1)}{a_{1}^{n+1}a_{2}^{b}} +e^{a_{2}p} \sum_{j=0}^{b-1} \binom{b-1}{j} \frac{(-p)^{b-1-j}\, \Gamma(j+n+2)}{(j+1)(a_{1}+a_{2})^{j+n+2}}\, {_{2}}F_{1}\left(1,\, j+n+2;\, j+2;\frac{a_{2}}{a_{1}+a_{2}}\right) \end{equation}

Not sure if the summation can be solved to simplify the solution but my hunch is that it can be with a bit more work.