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I am not able to solve the following sum. Can you please provide any hints ? $$ \sum_{y=1}^\infty {}_1F_1(1-y;2;-\pi\lambda c) \frac{\lambda^y}{y!} $$

Note that the 3rd parameter of the Confluent Hypergeometric Function includes parameter λ.

Thank you for your time

Harry Peter
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dioxen
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2 Answers2

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Integral method using the associated Laguerre polynomials with $\;x:=-\pi\lambda c$ \begin{align} \sum_{n=1}^\infty {}_1F_1(1-n;2;x) \frac{\lambda^n}{n!}&=\sum_{n=1}^\infty \left[1+\sum_{k=1}^\infty \frac{(1-n)_k}{(2)_k}\frac{x^k}{k!}\right] \frac{\lambda^n}{n!}\\ &=\sum_{n=1}^\infty \sum_{k=0}^{n-1} \frac{(n-1)!}{(n-k-1)!(k+1)!}\frac{(-x)^k}{k!} \frac{\lambda^n}{n!}\\ &=\sum_{m=0}^\infty \sum_{k=0}^m \frac{m!}{(m-k)!(k+1)!}\frac{(-x)^k}{k!} \frac{\lambda^{m+1}}{(m+1)!}\quad\text{for}\ m:=n-1\\ &=\sum_{m=0}^\infty \frac{ L_m^1(x)}{(m+1)!}\frac{\lambda^{m+1}}{m+1}\\ &=\int \sum_{m=0}^\infty \frac{ L_m^1(x)}{(m+1)!}\lambda^m\;d\lambda\\ &=\int \frac{e^{\lambda}J_1\bigl(2\sqrt{x\lambda}\bigr)}{\sqrt{x\lambda}}\;d\lambda\\ &=\int_0^{2\sqrt{x\lambda}} \frac{e^{u^2/(4x)}}x\;J_1(u)\;du\quad\quad\text{(using}\ u:=2\sqrt{x\lambda}\,\text{)}\\ \end{align} Since $\ \displaystyle L_m^1(x)= \sum_{k=0}^m \frac{(m+1)!}{(m-k)!\,(1+k)!\,k!}(-x)^k\ $ and $\ \displaystyle \sum_{m=0}^\infty \frac{L_m^1(x)}{(m+1)!}\lambda^m=\frac{e^{\lambda}J_1\bigl(2\sqrt{x\lambda}\bigr)}{\sqrt{x\lambda}}$
(formula $(5)$ and $(18)$ from the MathWorld link)

Initial method (returning alternative formulations...)

Let's expand the hypergeometric series using the Pochhammer symbols : \begin{align} \sum_{n=1}^\infty {}_1F_1(1-n;2;-\pi\lambda c) \frac{\lambda^n}{n!}&=\sum_{n=1}^\infty \left[1+\sum_{k=1}^\infty \frac{(1-n)_k}{(2)_k}\frac{(-\pi\lambda c)^k}{k!}\right] \frac{\lambda^n}{n!}\\ &=e^\lambda-1+\sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{(n-1)!}{(n-k-1)!(k+1)!}\frac{(\pi\lambda c)^k}{k!} \frac{\lambda^n}{n!}\\ &=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \sum_{n=k+1}^\infty\frac{\lambda^n}{n\;(n-k-1)!}\\ &=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \sum_{n=k+1}^\infty\frac{\lambda^{n-1}}{(n-k-1)!}d\lambda\\ &=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \lambda^k\sum_{m=0}^\infty\frac{\lambda^m}{m!}d\lambda\\ &=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \lambda^k e^{\lambda}\;d\lambda\\ &=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\gamma(k+1,-\lambda)}{k!(k+1)!}\\ \end{align}

(using the definition of the incomplete gamma function : $\ \displaystyle\gamma(a,x)=\int_0^xt^{a-1}e^{-t}\;dt\ $)

\begin{align} \sum_{n=1}^\infty {}_1F_1(1-n;2;-\pi\lambda c) \frac{\lambda^n}{n!}&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;k!\left(1-e^{\lambda}\sum_{m=0}^k\frac {(-\lambda)^m}{m!}\right)}{k!(k+1)!}\\ &=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\left(1-e^{\lambda}\sum_{m=0}^k\frac {(-\lambda)^m}{m!}\right)}{(k+1)!}\\ &=e^\lambda+\frac{e^{-\pi\lambda c}-1}{\pi\lambda c}+e^{\lambda}\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\sum_{m=0}^k\frac {(-\lambda)^m}{m!}}{(k+1)!}\\ \end{align}

At this point it would be interesting to know if : $$f(x,y):=\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$ may be simplified (I don't know...)

Raymond Manzoni
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  • Thank you Raymond for your swift reply. I have tried to use the same methodology for reaching to a final result. Yet, it seems that I get another sum of a different form, provided that, based on my recursions, I only seem to lower the factorial degree by constantly changing the variables of summation. Do I miss something? – dioxen May 21 '13 at 14:12
  • @dioxen: The idea was that, with one of the factorials removed, you could get something simpler like a geometric series to further simplify the result (possibly again changing the order of the sum over $k$ and $m$). But I am currently editing my answer to get a more direct way : using associated Laguerre polynomials. (At this point it looks like the integral of the expression $(18)$ i.e. of a Bessel $J_1$ function but it is not finished nor sure so...) – Raymond Manzoni May 21 '13 at 21:57
  • Dear Raymond, it seems that you haven't taken into account that x=g(λ) during the integration / differentiation with respect to λ? Because, this was the challenging fact in the above problem and the key difference from the result in here (http://math.stackexchange.com/questions/377220/sum-involving-the-hypergeometric-function-power-and-factorial-functions). Once again, thank you for your invaluable help. – dioxen May 22 '13 at 09:28
  • @dioxen : Wishing independence of these variables was in fact one of the reasons to use another variable $x$, I needed to integrate only the power of $\lambda$ ! Anyway once you computed $f(x,\lambda)$ you may always deduc any $f(a\lambda,\lambda)$. – Raymond Manzoni May 22 '13 at 10:03
  • @dioxen: Concerning the final $f(x,y)$ a little observation : $f(x,y)+f(y,x)=e^{x+y}+I_0(2\sqrt{xy})$ so that the case $x=y$ is easy : $f(x,x)=\frac 12\left(e^{2x}+I_0(2x)\right)$. Probably not very helpful in the general case (so that I'll keep that here instead of in your dedicated thread) but Alpha for example seems unaware of this... – Raymond Manzoni May 22 '13 at 23:13
  • In fact $\sum\limits_{k=0}^\infty\dfrac{x^k}{k!}\sum\limits_{m=0}^k\dfrac {y^m}{m!}$ belongs to $\Phi_3$ type series according to http://en.wikipedia.org/wiki/Humbert_series. – Harry Peter Aug 23 '16 at 12:27
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$\sum\limits_{y=1}^\infty{}_1F_1(1-y;2;-\pi\lambda c)\dfrac{\lambda^y}{y!}$

$=\sum\limits_{y=0}^\infty{}_1F_1(-y;2;-\pi\lambda c)\dfrac{\lambda^{y+1}}{(y+1)!}$

$=\sum\limits_{y=0}^\infty\sum\limits_{n=0}^y\dfrac{\pi^nc^n\lambda^{n+y+1}}{(2)_nn!(y-n)!(y+1)}$

$=\sum\limits_{n=0}^\infty\sum\limits_{y=n}^\infty\dfrac{\pi^nc^n\lambda^{n+y+1}}{(2)_nn!(y-n)!(y+1)}$

$=\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{\pi^nc^n\lambda^{2n+y+1}}{(2)_nn!y!(n+y+1)}$

$=\int_0^\lambda\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{\pi^nc^n\lambda^nx^{n+y}}{(2)_nn!y!}~dx$

$=\int_0^\lambda\sum\limits_{n=0}^\infty\dfrac{\pi^nc^n\lambda^nx^ne^x}{(2)_nn!}~dx$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}\pi^nc^n\lambda^nx^ke^x}{(2)_nk!}\right]_0^\lambda$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}\pi^nc^n\lambda^{n+k}e^\lambda}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^nc^n\lambda^n}{(2)_n}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^{n-k}\pi^nc^n\lambda^{n+k}e^\lambda}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^nc^n\lambda^n}{(n+1)!}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^{n+k}c^{n+k}\lambda^{n+2k}e^\lambda}{(2)_{n+k}k!}-\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}\pi^{n-1}c^{n-1}\lambda^{n-1}}{n!}$

$=e^\lambda\Phi_3(1,2;-\pi c\lambda,\pi c\lambda^2)+\dfrac{e^{-\pi c\lambda}-1}{\pi c\lambda}$ (according to http://en.wikipedia.org/wiki/Humbert_series)

Harry Peter
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