I feel like I'm losing my mind...
Letting $c$ be a positive real constant, consider the function $f\left(x\right)=e^{-cx^{2}}$ . This is one of the nicest functions anyone can ask for. Not only is it integrable over the real line, but its power series:
$e^{-cx^{2}}=\sum_{n=0}^{\infty}\left(-c\right)^{n}\frac{x^{2n}}{n!}$
converges uniformly everywhere.
Now, consider its laplace transform: $F\left(s\right)=\int_{0}^{\infty}e^{-cx^{2}}e^{-sx}dx$ The integral couldn't be more well behaved. It is uniformly convergent for all $s$ in any bounded subset of $\mathbb{C}$ . Since everything is uniformly convergent, I can write $e^{-cx^{2}}$ as its power series in the integral, and freely interchange sum and integral:
$\int_{0}^{\infty}e^{-cx^{2}}e^{-sx}dx = \int_{0}^{\infty}\sum_{n=0}^{\infty}\left(-c\right)^{n}\frac{x^{2n}}{n!}e^{-sx}dx = \sum_{n=0}^{\infty}\frac{\left(-c\right)^{n}}{n!}\int_{0}^{\infty}x^{2n}e^{-sx}dx = \sum_{n=0}^{\infty}\frac{\left(2n\right)!}{n!}\frac{\left(-c\right)^{n}}{s^{2n+1}}$
where I have used the fact that:
$\int_{0}^{\infty}x^{2n}e^{-sx}dx=\frac{\left(2n\right)!}{s^{2n+1}}$ Now comes the paradox: $\sum_{n=0}^{\infty}\frac{\left(2n\right)!}{n!}\frac{\left(-c\right)^{n}}{s^{2n+1}}$ diverges for all $s\in\mathbb{C}$!
And yet, as a quick consultation with Dr. Wolfram Alpha tells us:
$\mathcal{L}\left\{ e^{-cx^{2}}\right\} \left(s\right)=\frac{1}{2}\sqrt{\frac{\pi}{c}}e^{\frac{s^{2}}{4c}}\textrm{erfc}\left(\frac{s}{2\sqrt{c}}\right)$ where $\textrm{erfc}$ is the complementary error function:$\textrm{erfc}\left(z\right)=\frac{2}{\sqrt{\pi}}\int_{z}^{\infty}e^{-t^{2}}dt$
Now, my questions:
1) What is going on here? Why does the interchange of sum and integral blow up in my face? Is there some arcane detail of series and integral convergence that I am overlooking which makes the above sum and integral interchange invalid?
2) I am working with a family of related functions defined as follows: $\mathcal{G}_{p}\left(x\right)=\left(\frac{3}{2}\right)^{p}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\frac{x^{2n}}{\left(n+\frac{3}{2}\right)^{p}}$ where $p$ is a non-negative integer. (Note that $\mathcal{G}_{0}\left(x\right)=e^{-x^{2}}$ ) All I want is to compute the integral:
$\int_{0}^{\infty}\mathcal{G}_{p}\left(x\right)e^{-x}dx$ However, if I plug in the above series definition, then, just as with $e^{-cx^{2}}$ , I end up obtaining divergent gobbledygook.
As such, there seems to me only two ways to proceed:
i. Understand what goes wrong with the term-by-term laplace transform of $e^{-cx^{2}}$ so as to be able to evaluate the above integral in a manner that yields a convergent value for the integral.
ii. Sum the series defining $\mathcal{G}_{p}\left(x\right)$ so as to obtain a function for which the above integral can be transformed into a convergent expression, similar to how:
$\mathcal{L}\left\{ e^{-cx^{2}}\right\} \left(s\right)=\frac{1}{2}\sqrt{\frac{\pi}{c}}e^{\frac{s^{2}}{4c}}\textrm{erfc}\left(\frac{s}{2\sqrt{c}}\right)$
To give one more piece of pertinent background information, I obtained $\mathcal{G}_{p}\left(x\right)$ by applying the integral operator: $\Upsilon\left\{ f\right\} \left(x\right)\overset{\textrm{def}}{=}\frac{1}{x}\int_{0}^{x}f\left(t\right)dt$ to $e^{-x^{2}}$ $p$ times.
Thank you in advance for your assistance.
A necessary condition for a series of functions
$$ \sum_{k=0}^\infty f_k$$
to converge uniformly on $A \subseteq \mathbb R$ is
$$ \lim_{k \to+\infty} \sup_{x \in A} |f_k(x)| = 0.$$
In your case (the exponential series), you have
$$\sup_{x \in (0,+\infty)} \left| \frac{x^{2k}}{k!}(-c)^k \right| = +\infty.$$
for all $k$. Therefore the series defining $e^{-cx^2}$ does not converge uniformly on $(0,+\infty)$ and that explains why you couldn't exchange the series with the integral ;)