I first posed this question when I was a first year student. I came up with some ad hoc arguments as to why the result is true (a bit of numerical experimentation), but never had a proof. I forgot about it until the other day, and thought it was still an interesting question. First, some motivation.


Take the expression


$$\sum\_{n=0}^\infty \frac{x^n}{n!}$$


which equals its own derivative, that much is straight forward. The $x^n$ gets sent to $n x^{n-1}$ and the factorial kills the $n$, and the same expression emerges. 


We can do the same thing with the integral.


$$\int\_{-\infty}^\infty \frac{x^y}{\Gamma(y+1)}\,dy$$


formally satisfies being its own derivative--albeit the expression does not converge (which is a shame). But if you act like it converges, it should converge to $e^x$ because it's its own derivative.


This led me to considering a more suitable case, where the integral actually converges, and considering if the two functions agree. For example,


$$F(x) = \int\_{-\infty}^{\infty}e^{-y^2} \frac{x^y}{\Gamma(y+1)}\,dy$$


defines a $C^{\infty}$ function in $x$ for $x \in \mathbb{R}^+$. This function satisfies $F'(x) = e^{-1}F(\frac{x}{e^2})$. 


If we take


$$G(x) = \sum\_{n=0}^\infty e^{-n^2}\frac{x^n}{n!}$$


then equally so, $G'(x) = e^{-1}G(\frac{x}{e^2})$. 



> 
> Does $G = F$?
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> 
> 


I thought of this in a more general setting though. I wondered if it happened for $e^{-y^2}$, does it work for other functions?


It may be useful to explain the ad hoc arguments I had when I was younger. These arguments were a summation of identities which are satisfied by both expressions, which seems implausible if they weren't equal in some sense. I'll explain them in the obtuse language I used when I was a first year, as it is the best way I can think of framing it. Let $\mathcal{I}$ and $\mathcal{S}$ denote the following operators,


$$\mathcal{I} \{f\} (x) = \int\_{-\infty}^\infty f(y)\frac{x^y}{\Gamma(y+1)}\,dy$$


$$\mathcal{S} \{f\} (x) = \sum\_{n=0}^\infty f(n) \frac{x^n}{n!}$$


Also, let $Qf(x) = xf(x)$ and $Tf(x) = f(x+1)$. Then $\mathcal{S}$ and $\mathcal{I}$ behave identically under actions of the two operators. The following similar identities present evidence as to why these two operators may be equivalent in a good enough scenario.


$$\mathcal{I}\{Tf\}(x) = \frac{d}{dx}\mathcal{I} \{f\} (x)$$


$$\mathcal{S} \{Tf\} (x) = \frac{d}{dx}\mathcal{S} \{f\} (x)$$


$$\mathcal{I}\{Qf\}(x) = Q\mathcal{I}\{Tf\}(x) = Q\frac{d}{dx}\mathcal{I} \{f\} (x)$$
$$\mathcal{S} \{Qf\} (x) = Q\mathcal{S} \{Tf\} (x) = Q\frac{d}{dx}\mathcal{S} \{f\} (x)$$


From these we also get


$$Q \mathcal{I}\{f\}(x) = \mathcal{I}\{QT^{-1}f\}(x)$$
$$Q \mathcal{S}\{f\}(x) = \mathcal{S}\{QT^{-1}f\}(x)$$


From all this the question is simple,



> 
> Are these two operators ever equivalent, if so when?
> 
> 
> 


Or phrased more poetically



> 
> Is the continuous Taylor series no different than the discrete Taylor series?
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> 



The continuous Taylor series; are they just Taylor series?
No. In fact I will show the map from $1$-periodic signed measures $\mu$ to $$F\_\mu(x) = \int\_{-\infty}^\infty \mu(y) e^{-y^2}\frac{x^y}{\Gamma(y+1)}dy$$ is injective for "reasonable" signed measures $\mu$.


Indeed, consider what happens to the following expression as $y\_0$ goes to $\infty$ while the residue of $y\_0$ modulo $1$ remains fixed: 


$$ \Gamma(y\_0+1) e^{-y\_0^2 - y\_0 \operatorname{dlog} \Gamma(y\_0+1) } F\_{\mu} ( e^{ 2 y\_0 + \operatorname{dlog} \Gamma(y\_0+1)} ) $$


$$ = \int\_{-\infty}^{\infty} \mu(y) e^{-y^2 + 2 y y\_0 - y\_0^2} \frac{ \Gamma(y\_0+1) e^{ y \operatorname{dlog} \Gamma(y\_0+1)- y\_0 \operatorname{dlog} \Gamma(y\_0+1)}}{\Gamma(y+1)} dy$$


$$ = \int\_{\infty}^{\infty} \mu(y+y\_0) e^{-y^2} \frac{ \Gamma(y\_0+1) e^{ y \operatorname{dlog} \Gamma(y\_0+1)}}{\Gamma(y+y\_0+1)} dy$$


Because $\Gamma(y+1)$ is log-convex and its slope is roughly $\log y$, the exponential-linear approximation to the slope gets better as $y\_0$ goes to $\infty$, and so this


$$ \approx \int\_{-\infty}^{\infty} \mu(y+y\_0) e^{-y^2} dy $$


uniquely determines the convolution of $\mu$ with the Gaussian. Now because none of the Fourier transform coefficients of the Gaussian vanish, we can extract all the Fourier series coefficients of $\mu$ from this, which under mild analytic conditions extracts $\mu$. 


In particular, for your example $F(x)$ we see that $F(e^{2 y\_0 + \operatorname{dlog} \Gamma(y\_0+1)})$ is asymptotic to a constant times $\frac{ e^{y\_0^2 + y\_0 \operatorname{dlog} \Gamma(y\_0+1) }}{ \Gamma(y\_0+1)}$ while this is not true for your Taylor series $G(x)$.


