6

I am finding some trouble in calculating the following sum involving the hypergeometric function, power and factorial functions. $$ \sum_{y=1}^\infty \mathrm{e}^z \cdot {}_1F_1\left(1-y;2;-z\right) \frac{\mu^y}{y!} \mathrm{e}^{-\mu} $$

Could you please provide me some hints to solve this sum? Thank you in advance

Harry Peter
  • 7,819
dioxen
  • 510

2 Answers2

4
  1. Multiply your sum by $e^{\mu-z}$ and differentiate with respect to $\mu$ to get $$\sum_{n=0}^{\infty} {}_1F_1(-n,2,-z)\frac{\mu^n}{n!}\tag{1}$$
  2. Use the (I hope, accidentally) unaccepted answers to your question here to express $_1F_1$ in the sum as generalized Laguerre polynomials: $$ \frac{_1F_1(-n,2,-z)}{n!}=\frac{L_n^{(1)}(-z)}{(1+1)_n}.\tag{2}$$
  3. Use (2) and the generating function of Laguerre polynomials (formula 18.12.14) to sum up the series (1) to $$ \left(-\mu z\right)^{-1/2}e^{\mu}J_1(2\sqrt{-\mu z})=\left(\mu z\right)^{-1/2}e^{\mu}I_1(2\sqrt{\mu z})$$
  4. Integrate back with respect to $\mu$ and multiply the result by $e^{z-\mu}$. The final result is $$ e^{z-\mu}\int_0^{\mu}\left(\nu z\right)^{-1/2}e^{\nu}I_1(2\sqrt{\nu z})\,d\nu=z^{-1}e^{z-\mu}\int_0^{2\sqrt{\mu z}}e^{{x^2}/{4z}}I_1(x)\,dx.$$ I don't think the last integral can be expressed in terms of elementary or reasonably simple special functions.
Community
  • 1
Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
  • I was heading there but put it away for later. It seems you got where I was heading. (+1) – Ron Gordon Apr 30 '13 at 15:35
  • Thank you O.L. for the constant help provided. My question arised in continuation of the derivation here http://math.stackexchange.com/questions/369364/expected-value-of-a-poisson-sum-of-confluent-hypergeometric-functions where I didn't seem to underestand the derivation in the last line. Any ideas from combining your answer with this one? Thank you again for the support. – dioxen May 01 '13 at 08:50
  • O.L. and @Ron : I didn't know this thread but answered a more recent one with the same question and result (after some more exotic attempts). Glad to meet you both here even if late ! :-) – Raymond Manzoni May 22 '13 at 06:37
  • or at least 'nearly' the same question ($-n\to 1-n$)... – Raymond Manzoni May 22 '13 at 06:49
1

$\sum\limits_{y=1}^\infty e^z\cdot{}_1F_1(1-y;2;-z)\dfrac{\mu^y}{y!}e^{-\mu}$

$=e^{z-\mu}\sum\limits_{y=0}^\infty{}_1F_1(-y;2;-z)\dfrac{\mu^{y+1}}{(y+1)!}$

$=e^{z-\mu}\sum\limits_{y=0}^\infty\sum\limits_{n=0}^y\dfrac{\mu^{y+1}z^n}{(2)_nn!(y-n)!(y+1)}$

$=e^{z-\mu}\sum\limits_{n=0}^\infty\sum\limits_{y=n}^\infty\dfrac{\mu^{y+1}z^n}{(2)_nn!(y-n)!(y+1)}$

$=e^{z-\mu}\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{\mu^{n+y+1}z^n}{(2)_nn!y!(n+y+1)}$

$=e^{z-\mu}\int_0^\mu\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{x^{n+y}z^n}{(2)_nn!y!}~dx$

$=e^{z-\mu}\int_0^\mu\sum\limits_{n=0}^\infty\dfrac{x^ne^xz^n}{(2)_nn!}~dx$

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

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}\mu^kz^ne^z}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nz^ne^{z-\mu}}{(2)_n}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^{n-k}\mu^kz^ne^z}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nz^ne^{z-\mu}}{(n+1)!}$

$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\mu^kz^{n+k}e^z}{(2)_{n+k}k!}-\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}z^{n-1}e^{z-\mu}}{n!}$

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

Harry Peter
  • 7,819