A curious connection : What's the function $f(x)$? Hello I was wondering what was the function $f$ defines like this :

Let $f(x)$ a continuous and differentiable function such that :
  $$f(x)=\sum_{k=0}^{\infty}\frac{f'(k)}{k!}(-x)^k$$

In fact can't solve it but it makes a connection between Ramanujan's Master theorem and Frullani's integral via the Fundamental theorem of calculus I explain  :
We have :
$$\int_{0}^{\infty}x^{-s-1}f(x)dx=\Gamma{(-s)}f'(s)$$
Or :
$$\int_{0}^{\infty}\frac{x^{-s-1}f(x)}{\Gamma{(-s)}}dx=f'(s)$$
Now we use the Fundamental theorem of calculus to get :
$$\int_{0}^{s}\int_{0}^{\infty}\frac{x^{-s-1}f(x)}{\Gamma{(-s)}}dxds=f(s)-f(0)$$
Now we take the limit to get :
$$\lim_{s\to\infty}\int_{0}^{s}\int_{0}^{\infty}\frac{x^{-s-1}f(x)}{\Gamma{(-s)}}dxds=f(\infty)-f(0)$$
Wich is equal to:
$$\int_{0}^{\infty}\frac{f(ax)-f(bx)}{ln(\frac{a}{b})x}$$
So my question is what is the function $f(x)$ , there exists a closed form to this ,is it trivial or not ? 
Thanks 
Ps:I know it's not very rigorous but I think it's interesting 
 A: Not a full answer, but a set of equations to exploit.
Let 
$$
g(x) = f'(x)=\sum_{k=1}^{\infty}\frac{k f'(k)}{k!}(-1)^k x^{k-1}= \sum_{k=0}^{\infty}\frac{g(k+1)}{k!}(-1)^{k+1} x^{k}
$$
But also, via Taylor expansion
$$
g(x) = \sum_{k=0}^{\infty}\frac{g^{(k)}(0)}{k!} x^{k}
$$
So we obtain the set of equations, for all $k=0 \cdots \infty$:
$$
g^{(k)}(0) = {g(k+1)}(-1)^{k+1}
$$
The first few are:
$$
g(0) = - g(1)\\
g'(0) = g(2)\\
g''(0) = - g(3)\\
g^{(3)}(0) = g(4)\\
\cdots
$$
Can we proceed from here?
A: Hint: There is a relationship of $f(x)$ with the Stirling numbers of the second kind.

Using the ansatz $f(x)=\sum_{j=0}^{\infty}a_j\frac{x^j}{j!}$ we obtain
  \begin{align*}
\color{blue}{f(x)}&=\sum_{k=0}^\infty \frac{f^\prime(k)}{k!}(-x)^k\\
&=\sum_{k=0}^\infty\frac{d}{du}\left.\left(\sum_{j=0}^\infty a_j\frac{u^j}{j!}\right)\right|_{u=k}\frac{(-x)^k}{k!}\\
&=\sum_{k=0}^\infty\left.\left(\sum_{j=1}^\infty a_j\frac{u^{j-1}}{(j-1)!}\right)\right|_{u=k}\frac{(-x)^k}{k!}\\
&=\sum_{k=0}^\infty\left(\sum_{j=0}^\infty a_{j+1}\frac{k^j}{j!}\right)\frac{(-x)^k}{k!}\\
&=\sum_{j=0}^\infty\frac{a_{j+1}}{j!}\sum_{k=0}^\infty\frac{k^j}{k!}(-x)^k\\
&=\sum_{j=0}^\infty\frac{a_{j+1}}{j!}e^{-x}\sum_{k=0}^j{j\brace k}(-x)^k\tag{1}\\
&=\sum_{j=0}^\infty \frac{a_{j+1}}{j!}[t^j]e^{-xe^t}\tag{2}\\
&=[t^0]e^{-xe^t}\sum_{j=0}^\infty a_{j+1}\frac{t^{-j}}{j!}\\
&\,\,\color{blue}{=[t^0]e^{-xe^t}\left(\left.f^{\prime}(x)\right|_{x=\frac{1}{t}}\right)}\tag{3}
\end{align*}
According to (3) the bivariate formal Laurent series $F(x,t)=e^{-xe^t}\sum_{j=0}^\infty a_{j+1}\frac{t^{-j}}{j!}$ might be useful to find a representation of $f(x)$.

Comment:


*

*In (1) we note that $\sum_{k=0}^\infty \frac{k^j}{k!}x^k$ admits a representation via Stirling numbers of the second kind times $e^x$.

*In (2) we use the bivariate generating function of the Stirling numbers of the second kind
\begin{align*}
e^{x(e^t-1)}&=\sum_{k=0}^\infty \frac{(e^t-1)^kx^k}{k!}\\
&=\sum_{k=0}^\infty\left(\sum_{j=k}^\infty {j\brace k} \frac{t^j}{j!}\right)x^k\\
&=\sum_{j=0}^\infty\left(\sum_{k=0}^j{j\brace k}x^k\right)\frac{t^j}{j!}\\
\end{align*}
from which we get by using $-x$ instead of $x$ and selecting the coefficient of $t^j$:
\begin{align*}
\frac{1}{j!}e^{-x}\sum_{k=0}^j{j\brace k}(-x)^k=[t^j]e^{-xe^t}
\end{align*}
A: Considering the operator
$$
\cal{L}(f) = \sum_{k=0}^{\infty}\frac{f'(k)}{k!}(-x)^k
$$
and choosing 
$$
f(x) = e^{-W(1) x}
$$
with $W(\cdot)$ the Lambert function, we have
$$
\cal{L}(f) =\mbox{ $-\frac{1}{W(1)}f$}
$$
because
$$
f(x) = 1-x W(1)+\frac 12 W(1)^2 x^2-\frac 16 W(1)^3 x^3+\cdots + \frac{W(1)^k}{k!}(-x)^k + \cdots\\
\frac{f(x)}{W(1)} = -1 + e^{-W(1)}x-\frac 12e^{-2W(1)}x^2+\frac 16 e^{-3W(1)}x^3+\cdots + \frac{e^{-kW(1)}(-x)^k}{k!}+\cdots
$$
with
$$
W(1)^k = e^{-k W(1)}
$$
I hope it helps.
