How to sum up this series or find a closed expression or argument?

Question

We know the following (Taylor expansion) $$\sum_{n=0}^\infty \frac{(-1)^n}{n!} (x^2)^n = e^{-x^2}.$$ Which is integrable in $x$ on $\mathbb{R}$. Nevertheless, imagine we instead have:

$$\sum_{n=0}^\infty \frac{(-1)^n}{n!} (x^2)^n \frac{\Gamma(1+n)}{\Gamma(1+a+n)},$$ where $a>0$ is some fixed value. Clearly, $$\frac{\Gamma(1+n)}{\Gamma(1+a+n)}\leq 1$$ so one should expect that this factor does not contibute to the sum and that the remaining function in $x$ is still integrable on $\mathbb{R}$.

Of course, one can say: $$\sum_{n=0}^\infty \frac{(-1)^n}{n!} (x^2)^n \frac{\Gamma(1+n)}{\Gamma(1+a+n)}\leq e^{x^2}.$$ Using the inequality, but this implies destroying the $-1$ which is crucial.

Does any one have an idea or argument that can help to either "keep" the $-1$ or to conclude that the remaining function is integrable on $\mathbb{R}$?

Thanks a lot for any ideas you may have! :)

Answer 1

For $a\in\mathbb N$ and $n\in\mathbb N$, we have

$$\frac{\Gamma(1+n)}{\Gamma(1+a+n)}=\frac{n!}{(n+a)!}$$

Notice that

$$\frac{x^n}{n!}\frac{\Gamma(1+n)}{\Gamma(1+a+n)}=\frac{x^n}{n!}\frac{n!}{(n+a)!}=\frac{x^n}{(n+a)!}$$

$$\sum_{n\ge0}\frac{x^n}{n!}\frac{\Gamma(1+n)}{\Gamma(1+a+n)}=\sum_{n\ge0}\frac{x^n}{(n+a)!}=\frac1{x^a}\left(e^x-\sum_{n=0}^{a-1}\frac{x^n}{n!}\right)$$

And one can set bounds on this by seeing that

$$\frac{\Gamma(1+n)}{\Gamma(1+\lceil a\rceil+n)}\le\frac{\Gamma(1+n)}{\Gamma(1+a+n)}\le\frac{\Gamma(1+n)}{\Gamma(1+\lfloor a\rfloor+n)}$$

So your series/function is bounded by

$$\left|\frac1{x^{\lceil a\rceil}}\left(e^x-\sum_{n=0}^{\lceil a\rceil-1}\frac{x^n}{n!}\right)\right|\le|S|\le\left|\frac1{x^{\lfloor a\rfloor}}\left(e^x-\sum_{n=0}^{\lfloor a\rfloor-1}\frac{x^n}{n!}\right)\right|$$

For $x>0$.

And finally let $x\to-x^2$

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(Note that the negative sign may cause some problems with the bounds on this.)

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!}x^{2n}\, {\Gamma\pars{1 + n} \over \Gamma\pars{1 + a + n}} = {1 \over \Gamma\pars{a}}\sum_{n = 0}^{\infty}{\pars{-x^{2}}^{n} \over n!}\, {\Gamma\pars{a}\Gamma\pars{1 + n} \over \Gamma\pars{1 + a + n}} \\[5mm] = &\ {1 \over \Gamma\pars{a}}\sum_{n = 0}^{\infty} {\pars{-x^{2}}^{n} \over n!}\int_{0}^{1}t^{n}\pars{1 - t}^{a - 1}\,\dd t = {1 \over \Gamma\pars{a}}\int_{0}^{1}\pars{1 - t}^{a - 1}\sum_{n = 0}^{\infty} {\pars{-x^{2}t}^{n} \over n!}\,\dd t \\[5mm] = &\ {1 \over \Gamma\pars{a}}\int_{0}^{1}\pars{1 - t}^{a - 1}\expo{-x^{2}t}\,\dd t = {\expo{-x^{2}} \over \Gamma\pars{a}}\int_{0}^{1}t^{a - 1}\expo{x^{2}t}\,\dd t \\[5mm] & = \bbx{\ds{% {1 \over \Gamma\pars{a}}\,{\expo{-x^{2}} \over x^{2a}}\int_{0}^{x^{2}}t^{a - 1}\expo{t}\,\dd t}}\label{1}\tag{1} \end{align}

The above $\underline{\texttt{final}}$ result ( see expression \eqref{1} ) will involve the Gamma and the Incomplete Gamma functions but some care must be taken as related to a $\ds{\pars{-1}^{\, a}}$-branch cut.