A Paradoxical Laplace Transform 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: Your Laplace transform integral can be integrated directly as the limit for $a\to \infty $ of
1)$F(s)=\int _{0}^{a}e^{-cx^{2} } e^{-sx}  dx=\frac{1}{\sqrt{c} } e^{\frac{s^{2} }{4c} } \int _{\frac{s}{2\sqrt{c} } }^{\sqrt{c} a+\frac{s}{2\sqrt{c} } }e^{-t^{2} }  dt=\frac{1}{2} \sqrt{\frac{\pi }{c} } e^{\frac{s^{2} }{4c} } \left\{\right. erf(\sqrt{c} a+\frac{s}{2\sqrt{c} } )-erf(\frac{s}{2\sqrt{c} } )\left. \right\}$
where we have used the substitution $t=\sqrt{c} x+\frac{s}{2\sqrt{c} } $ and where erf() is the error function.
Since $erf(\infty )=1$ we obtain for $a\to \infty $
2) $F(s)=\frac{1}{2} \sqrt{\frac{\pi }{c} } e^{\frac{s^{2} }{4c} } erfc(\frac{s}{2\sqrt{c} } )$ 
which is in accordance with your Dr. Wolfram consultation.
Your next steps essentially produce an asymptotic expansion of F(s) and the easiest way to get such an expansion is often to integrate by parts.
A simple way (amongst others) to show this is to introduce a new dummy variable $t=e^{-sx} $giving
3) $F(s)=\int _{0}^{a}e^{-cx^{2} } e^{-sx}  dx=\frac{1}{s} \int _{0}^{1}\exp (-c\frac{\ln ^{2} (t)}{s^{2} } ) dt$ 
Expanding exp() in its McLaurin series then gives
4) $F(s)=\int _{0}^{1}\sum _{n=0}^\infty (-c)^{n} \frac{1}{s^{2n+1}(n)!} \ln ^{2n} (t)dt$ 
Integrating by parts we have
5) $\int  \ln ^{2n} (t)dt=t\ln ^{2n} (t)-2n\int  \ln ^{2n-1} (t)dt$ 
Repeated use of this then gives
6)$ \int _{0}^{1} \ln ^{2n} (t)dt=t\sum _{p=0}^{2n} (-1)^{p} \frac{(2n)!}{(2n-p)!} \ln ^{2n-p} (t)\left|\begin{array}{c} {1} \\ {0} \end{array}\right. =(2n)!$
where we have made use of $t\cdot \ln (t)\to 0\, \, for\, \, t\to 0$.
Inserting (2n)! in 4) then reproduces your ''non-converging'' result
7) $F(s)=\sum _{n=0}^{\infty } \frac{1}{s^{2n+1} } \frac{(2n)!}{n!} (-c)^{n}$ 
All this is essentially the same as you have done but here done in a different way in order to underline that the above sum is an asymptotic expansion in 1/s of F(s). The difference between such an expansion and a "normal" Taylor expansion is that the convergence of the Taylor series is versus index n when $n\to \infty $ while the convergence of F(s) is versus s when $s\to \infty $ for some limited range of n, in which the expansion approximates F(s) for large s but never exactly equals F(s).
For more information I suggest you download the paper "Asymptotic Expansions", University of California, Davis, from the site https://www.math.ucdavis.edu/$\sim$hunter/m204/ch2.pdf . 
Here especially equation 2.4 deals with your expansion.
I also recommend the Wikipedia article about "Borel summation" which directly pertains to what you (and I) are doing.
A: 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 ;) 
