I would like to have a closed form or any nicer form for the following series:

$$\sum_{m=1}^{\infty}\,\frac{(-1)^m}{m!\Gamma (m+2)}\left(\frac{R}{2}\right)^{2m+1}\,x^{m}\sum_{j=1}^m\frac{(j-1)!}{(ax)^j}$$


$$\sum_{m=0}^{\infty}\,\frac{(-1)^m}{m!\Gamma (m+2)}\left(\frac{R}{2}\right)^{2m+1}\,x^{m+1}\sum_{j=1}^{m+1}\frac{(j-1)!}{(ax)^j}$$

with $R,a>0$.



So one needs to simplify $$\Sigma=\sum_{m=0}^{\infty}\,\frac{(-1)^m}{m!\Gamma (m+2)}\left(\frac{R}{2}\right)^{2m+1}\,x^{m+1}\sum_{j=1}^{m+1}\frac{(j-1)!}{(ax)^j}$$ First let's assume the inner sum: $$\sum_{j=1}^{m+1}\frac{(j-1)!}{(ax)^j}=\frac{1}{a x}\sum_{n=0}^{m}\frac{n!}{(ax)^n}=-(-1)^m (m+1)! e^{-a x} \Gamma (-m-1,-a x)-e^{-a x} \Gamma (0,-a x)$$ where $\Gamma (\cdot,\cdot)$ - is the incomplete Gamma function. So the series will split in to so series $\Sigma=\Sigma_1+\Sigma_2$: $$\Sigma_1=-\frac{xR}{2}e^{-ax}\sum_{m=0}^{\infty}\,\frac{\left(\frac{R^2x}{4}\right)^{m}}{m!}\,\Gamma (-m-1,-a x) \\ \Sigma_2=-\frac{xR}{2}e^{-a x} \Gamma (0,-a x)\sum_{m=0}^{\infty}\,\frac{(-1)^m}{m!\Gamma (m+2)}\left(\frac{R^2x}{4}\right)^{m}$$ Making the summation one gets: $$\Sigma_2=-\frac{xR}{2}e^{-a x} \Gamma (0,-a x)\frac{J_1\left(2\sqrt{\frac{R^2x}{4}}\right)}{\sqrt{\frac{R^2x}{4}}}=-\sqrt{x}J_1(R\sqrt{x})e^{-a x} \Gamma (0,-a x)$$ The computation of the first series is not that straightforward.
One can leave it like this: $$\Sigma=-\frac{xR}{2}e^{-ax}\sum_{m=0}^{\infty}\,\frac{\left(\frac{R^2x}{4}\right)^{m}}{m!}\,\Gamma (-m-1,-a x)-\sqrt{x}J_1(R\sqrt{x})e^{-a x} \Gamma (0,-a x)$$ Or use the integral representation of $\Gamma (\cdot,\cdot)$ (I'll skip this step): $$\Sigma_1=\frac{R}{2a}\int_0^{\infty } \frac{t e^{-t}}{t-a x} \, \mathrm dt \sum_{m=0}^{\infty}\,\frac{(-1)^m\left(\frac{R^2t}{4a}\right)^{m}}{m!\Gamma (m)}=-\frac{R^2}{4a^{\frac{3}{2}}}\int_0^{\infty }\frac{t \ J_1\left(\frac{R }{\sqrt{a}}\sqrt{t}\right)e^{-t}}{t-a x} \, \mathrm dt $$ And the last integral can be computed numerically or you can look through tables of integral of special function like Gradshteyn I.S., Ryzhik I.M. Tables of Integrals, Series, and Products.

  • $\begingroup$ Thanks..the final step you show is for $\Sigma$ and not $\Sigma_2$, right? $\endgroup$ – JFNJr Jul 30 '13 at 8:32
  • $\begingroup$ Ah.. yes, it should be $\Sigma_1$. I've corrected it. $\endgroup$ – Caran-d'Ache Jul 30 '13 at 9:54

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