1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\alpha,\beta$. $\endgroup$ – Jack D'Aurizio Jul 11 '16 at 14:11
  • $\begingroup$ 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. $\endgroup$ – Aaron Hendrickson Jul 11 '16 at 14:17
  • 1
    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jul 11 '16 at 17:14
  • $\begingroup$ 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. $\endgroup$ – Aaron Hendrickson Jul 11 '16 at 19:42
1
$\begingroup$

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

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

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

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

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

$=\sum\limits_{n=0}^\infty\dfrac{(-\alpha)_n\lambda^n}{(-\beta)_na^{n+1}}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-\alpha)_n\lambda^nx^ke^{-ax}}{(-\beta)_nk!a^{n-k+1}}\right]_0^{-x}$ (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)

$\endgroup$
1
$\begingroup$

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: $$ \sum_{m\geq 0}\frac{(-a)^m z^{m+1}}{(m+1)!}\phantom{}_2 F_2\left(1+m,-\alpha;2+m,-\beta;\lambda z\right).$$

$\endgroup$

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.