# Closed-form Solution for series involving incomplete Gamma Function

I am working on a solution for an intgeral that leads to a series that I am stuck at. Below is what I have done and how I got to the final series. Any ideas on how to solve the series at the end?

\begin{equation} \int_{-\infty}^{x} e^{ax} \, {_{1}}F_{1}(-\alpha;-\beta;-\lambda x) \ \mathrm{d}x \quad \text{for} \, x\leq 0 \end{equation}

where, $a,\alpha,\beta,\lambda>0$.

Substitute $u=-ax\to$ $\mathrm{d}u = -a \ \mathrm{d}x$

\begin{equation} a^{-1}\int_{-u/a}^{\infty} e^{-u} \, {_{1}}F_{1}(-\alpha;-\beta;\tfrac{\lambda}{a} u) \ \mathrm{d}u \end{equation}

At this point I could not find closed form solution for the integral so I wrote the confluent hypergeometric function in its summation representation.

\begin{equation} a^{-1}\sum_{j=0}^{\infty}\frac{(-\alpha)_{j}}{(-\beta)_{j}\, j!} \left(\frac{\lambda}{a}\right)^{j}\int_{-u/a}^{\infty} e^{-u} \, u^{j} \ \mathrm{d}u \end{equation}

where, $(x)_{j}$ is the Pochhammer symbol.

Since $u=-ax$, and $x\leq 0$, this means $-u/a\geq 0$ and the integral is a upper-incomplete gamma function.

\begin{equation} a^{-1}\sum_{j=0}^{\infty}\frac{(-\alpha)_{j}}{(-\beta)_{j}\, j!} \left(\frac{\lambda}{a}\right)^{j}\ \Gamma(j+1,x) \end{equation}

I do not know how to get past this point. An ideas on how to solve the sum?

From wolframalpha I found the following definition that could also prove to be useful:

\begin{equation} \gamma(m,x) = m^{-1} x^{m} \, {_{1}}F_{1}(m; m+1; -x) \end{equation}

where, $\Gamma(m) = \Gamma(m,x)+\gamma(m,x)$.

UPDATE

If $\alpha$ and $\beta$ are integers then a solution can be found through integration by parts. The solution is:

\begin{equation} a^{-1}\sum_{k=0}^{\alpha} \left(\frac{\lambda}{a}\right)^{k} e^{ax}\, {_{1}}F_{1}(-\alpha+k;-\beta+k,-\lambda x) \end{equation}

I am looking for the case where $\alpha$ and $\beta$ are not integers.

• If $\alpha,\beta\in\mathbb{N}$, your hypergeometric function is just a polynomial and the closed form is straightforward to compute. It shouldn't be terribly difficult to generalize such closed form for non-integer values of $\alpha,\beta$. – Jack D'Aurizio Jul 11 '16 at 14:11
• Jack, I have solved this problem for the case where $\alpha$ and $\beta$ are integers using ibp where $u={_{1}}F_{1}(-\alpha;-\beta,-\lambda x)$ and $\mathrm{d}v = e^{ax}$. I could include that answer in the post but the answer is still a series as a result of using ibp. And i'm not sure how it will help me get to the general solution. – Aaron Hendrickson Jul 11 '16 at 14:17
• I carried on some computations for the general case, and I am left with a hypergeometric function as a main term and a series of $\phantom{}_{2} F_2$ functions as secondary term, that does not seem to simplify further. I hope that helps. – Jack D'Aurizio Jul 11 '16 at 17:14
• Thanks Jack. I will take a look at it. It may be that there is no closed form solution. If no better answer arises I will accept what you posted. – Aaron Hendrickson Jul 11 '16 at 19:42

## 2 Answers

$\int_{-\infty}^xe^{ax}{_1}F_1(-\alpha;-\beta;-\lambda x)~dx$

$=\int_\infty^{-x}e^{-ax}{_1}F_1(-\alpha;-\beta;\lambda x)~d(-x)$

$=\int_{-x}^\infty e^{-ax}{_1}F_1(-\alpha;-\beta;\lambda x)~dx$

$=\int_{-x}^\infty\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^nx^ne^{-ax}}{(-\beta)_nn!}~dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^nx^ne^{-ax}}{(-\beta)_nn!}~dx-\int_0^{-x}\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^nx^ne^{-ax}}{(-\beta)_nn!}~dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^n}{(-\beta)_na^{n+1}}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-\alpha)_n\lambda^nx^ke^{-ax}}{(-\beta)_nk!a^{n-k+1}}\right]_0^{-x}$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^n}{(-\beta)_na^{n+1}}-\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^n}{(-\beta)_na^{n+1}}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-\alpha)_n\lambda^n(-1)^kx^ke^{ax}}{(-\beta)_nk!a^{n-k+1}}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-\alpha)_n\lambda^n(-1)^kx^ke^{ax}}{(-\beta)_nk!a^{n-k+1}}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_{n+k}\lambda^{n+k}(-1)^kx^ke^{ax}}{(-\beta)_{n+k}k!a^{n+1}}$

$=\dfrac{e^{ax}}{a}\Phi_1\left(-\alpha,1,-\beta;\dfrac{\lambda}{a},-\lambda x\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)

Let $z=-x$. We want to compute:

$$\int_{z}^{+\infty} e^{-at}\phantom{}_1 F_1(-\alpha,-\beta,\lambda t)\,dt=\sum_{n\geq 0}\frac{\lambda^n(-\alpha)_n}{n!(-\beta)_n}\int_{z}^{+\infty}e^{-at}t^n\,dt$$ where: $$\begin{eqnarray*} \int_{z}^{+\infty}e^{-at}t^n\,dt &=& \frac{n!}{a^{n+1}}-\int_{0}^{z}\sum_{m\geq 0}\frac{(-a)^m}{m!}\,t^{m+n}\,dt\\&=&\frac{n!}{a^{n+1}}-\sum_{m\geq 0}\frac{(-a)^m z^{m+n+1}}{(m+n+1)m!}\end{eqnarray*}$$ gives: $$\int_{z}^{+\infty} e^{-at}\phantom{}_1 F_1(-\alpha,-\beta,\lambda t)\,dt=\sum_{n\geq 0}\frac{\lambda^n (-\alpha)_n}{a^{n+1}(-\beta)_n}-\sum_{n\geq 0}\sum_{m\geq 0}\frac{\lambda^n(-\alpha)_n(-a)^m z^{m+n+1}}{m!n!(-\beta)_n(m+n+1)}$$ where the first series equals $\frac{1}{a}$ times $\phantom{}_2 F_1\left(1,-\alpha;-\beta;\frac{\lambda}{a}\right)$ and the second double series can be put in the following form by switching the order of summation: $$\sum_{m\geq 0}\frac{(-a)^m z^{m+1}}{(m+1)!}\phantom{}_2 F_2\left(1+m,-\alpha;2+m,-\beta;\lambda z\right).$$