Is there a neat closed form expression for $\frac{x^2}{\beta} \sum_{n=0}^{\infty} \frac{(-x/\beta)^n}{(n+2)! (n+1)}$?

Question

The original aim is to solve a differential equation of the form $$y_2(x) = \frac{1-exp(-\frac{x}{\beta})}{x}.$$ I tried a Taylor series expansion and got $$y_2(x) = \frac{1}{\beta}\sum_{n=0}^{\infty} \frac{(-x/\beta)^n}{(n+1)!\ }.$$ Then I tried integrating this expression twice to arrive at the form $$\frac{x^2}{\beta} \sum_{n=0}^{\infty} \frac{(-x/\beta)^n}{(n+2)! (n+1)}$$ I'd like to get a nice closed form expression of this term, if it exists.

Answer 1

Wolfy says that

$\sum_{n=0}^{\infty} \dfrac{x^{n + 2}}{(n + 1) (n + 2)!} \\\quad = - \gamma x + x - e^x - x \log(-x) - x Γ(0, -x) + 1 $

where $Γ(0, -x)$ is the incomplete gamma function.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.{x^{2} \over \beta} \sum_{n = 0}^{\infty} {\pars{-x/\beta}^{n} \over \pars{n + 2}!\pars{n + 1}} \,\right\vert_{\ \Re\pars{x/\beta}\ >\ 0}} = {x^{2} \over \beta}\sum_{n = 2}^{\infty} {\pars{-x/\beta}^{n - 2} \over n!\pars{n - 1}} \\[5mm] = &\ {x^{2} \over \beta}\,{\beta^{2} \over x^{2}} \sum_{n = 2}^{\infty} {\pars{-x/\beta}^{n} \over n!}\int_{0}^{1}t^{n - 2}\,\dd t = \beta\int_{0}^{1}\sum_{n = 2}^{\infty} {\pars{-xt/\beta}^{n} \over n!}\,{\dd t \over t^{2}} \\[5mm] = &\ \beta\int_{0}^{1}\pars{\expo{-xt/\beta} - 1 + {x \over \beta}\,t}\,{\dd t \over t^{2}} = x\int_{0}^{x/\beta}{\expo{-t} - 1 + t \over t^{2}}\,\dd t \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\, & x\bracks{\left.-\,{\expo{-t} - 1 + t \over t}\,\right\vert_{\ 0}^{\ x/\beta} - \int_{0}^{x/\beta}\pars{-\,{1 \over t}} \pars{-\expo{-t} + 1}\,\dd t} \\[8mm] = &\ -\beta\expo{-x/\beta} + \beta - x \\[2mm] &\ - x\lim_{\Lambda \to \infty} \bracks{\int_{0}^{\Lambda} {\expo{-t} - 1 \over t}\,\dd t -\ \underbrace{\int_{x/\beta}^{\Lambda} t^{\color{red}{0} - 1}\expo{-t} \,\dd t} _{\ds{\stackrel{\mrm{as}\ \Lambda\ \to\ \infty}{\LARGE\to}\ \Gamma\pars{0,{x \over \beta}}}}\ +\ \int_{x/\beta}^{\Lambda}{\dd t \over t}} \end{align}

$\ds{\Gamma\pars{a,z}}$ is the Incomplete Gamma Function.

Then, \begin{align} &\bbox[10px,#ffd]{\left.{x^{2} \over \beta} \sum_{n = 0}^{\infty} {\pars{-x/\beta}^{n} \over \pars{n + 2}!\pars{n + 1}} \,\right\vert_{\ \Re\pars{x/\beta}\ >\ 0}} \\[5mm] = &\ -\beta\expo{-x/\beta} + \beta - x + x\,\Gamma\pars{0,{x \over \beta}} + x\ \overbrace{\int_{0}^{\infty}\ln\pars{t}\expo{-t}\dd t} ^{\ds{=\ \gamma}} \\[2mm] &\ - x\ \underbrace{\lim_{\Lambda \to \infty}\bracks{% \ln\pars{\Lambda}\pars{\expo{-\Lambda} - 1} + \ln\pars{\Lambda \over x/\beta}}} _{\ds{-\ln\pars{x \over \beta}}} \end{align}

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$$ \implies \bbx{-x + \gamma x + \beta - \beta\expo{-x/\beta} + x\,\Gamma\pars{0,{x \over \beta}} + x\ln\pars{x \over \beta}} $$

$\ds{\gamma}$ is the Euler-Mascheroni Constant.