How to simplify $\sum_{k=0}^\infty \frac{(-1)^ka^{2k+1}}{(2k-1)(2k+1)!}$?

Question

I think that this sum can be simplified so that there's no factorial in the denominator and no exponential function in the numerator. But how can I do this? $a$ - is a constant. Could anybody show me step by step solution? $$\sum_{k=0}^\infty \frac{(-1)^ka^{2k+1}}{(2k-1)(2k+1)!}$$

Answer 1

$$\sum_{k\geq 0}\frac{(-1)^k a^{2k+1}}{(2k-1)(2k+1)!} = -a-a^2\sum_{j\geq 0}\frac{(-1)^j a^{2j+1}}{(2j+1)(2j+3)!}=-a-a^2\int_{0}^{a}\frac{x-\sin(x)}{x^3}\,dx $$ equals $$ -\frac{1}{2}\left(\sin(a)+a\cos(a)+a^2\text{Si}(a)\right) $$ where $\text{Si}(a)$ is the sine integral. The partial sums of this series have no particular closed form I am aware of.

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[15px,#ffc]{\sum_{k = 0}^{\infty}{\pars{-1}^{k}a^{2k + 1} \over \pars{2k - 1}\pars{2k + 1}!}} = -a - \ic\sum_{k = 1}^{\infty}{\ic^{2k + 1}a^{2k + 1} \over \bracks{\pars{2k + 1} - 2}\pars{2k + 1}!} \\[3mm] = &\ -a - \ic\sum_{k = 3}^{\infty} {\ic^{k}a^{k} \over \pars{k - 2}k!}\,{1^{k} - \pars{-1}^{k} \over 2} = -a + \Im\sum_{k = 3}^{\infty} {\pars{\ic a}^{k} \over \pars{k - 2}k!} \\[3mm] = &\ -a + \Im\sum_{k = 3}^{\infty} {\pars{\ic a}^{k} \over k!}\int_{0}^{1}t^{k - 3}\,\dd t = -a + \Im\int_{0}^{1}\sum_{k = 3}^{\infty} {\pars{\ic at}^{k} \over k!}\,{\dd t \over t^{3}} \\[3mm] = &\ -a + \Im\int_{0}^{1}\pars{\expo{\ic at} - 1 - \ic at + {a^{2}t^{2} \over 2}}{\dd t \over t^{3}} = -a + \int_{0}^{1}{\sin\pars{at} - at \over t^{3}}\,\dd t \\[3mm] = &\ \bbox[15px,#ffc,border:1px groove navy]{-\,{a^{2}\,\mrm{Si}\pars{a} + \sin\pars{a} + a\cos\pars{a} \over 2}}\ \pars{\mbox{after integrating repeatedly by parts}} \end{align}

$\ds{\mrm{Si}}$ is the Sine Integral Function.