$e^x=\sum \limits_{k=0}^\infty \frac{x^k}{k!}$
We can write $e^x=\sum \limits_{k=0}^\infty \frac{x^k}{ \Gamma(k+1)}$
Where $\Gamma(x)$ is Gamma function
$\Gamma(k+1)=k\Gamma(k)$
$\frac{\Gamma(k+1)}{\Gamma(k)}=k$
$\frac{\Gamma(1)}{\Gamma(0)}=0$
$\frac{1}{\Gamma(0)}=0$
$\frac{\Gamma(-1+1)}{\Gamma(-1)}=\frac{\Gamma(0)}{\Gamma(-1)}=-1$
$\frac{1}{\Gamma(-1)}=\frac{-1}{\Gamma(0)}=-1.\frac{1}{\Gamma(0)}=0$
If we continue in that way, we get result
for $m $ is non-positive integer, $\frac{1}{\Gamma(m)}=0$
Thus we can write $e^x=\sum \limits_{k=-\infty}^\infty \frac{x^k}{ \Gamma(k+1)}$
Then we extended $e^x$ for n is integer (Equation 1): $e^x=\sum \limits_{k=-\infty}^\infty \frac{x^{k+n}}{ \Gamma(k+n+1)}$
$n \in Z $ {...,-2,-1,0,1,2,...}
It is possible to extend the defination to $z \in C$.
$f(x)=\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)}$
$\frac{d(f(x))}{dx}=\frac{d}{dx}(\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)})= \sum \limits_{k=-\infty}^\infty (k+z)\frac{x^{k+z-1}}{ (k+z)\Gamma(k+z)}= \sum \limits_{k=-\infty}^\infty \frac{x^{k+z-1}}{ \Gamma(k+z)}=\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)}=f(x)$
$\frac{d(f(x))}{dx}=f(x)$
$f(x)=c(z)e^x$
$c(z)e^x=\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)}$ According to Equation 1, $c(z) = 1$ for $z \in Z$ but I noticed I need to find what is $c(z)$ for $z \in C$. (Thanks to Norbert for his contribution)
After that we can find the result:
$\frac{\partial(c(z)e^x)}{\partial z}=\sum \limits_{k=-\infty}^\infty \frac{\partial}{\partial z}(\frac{x^{k+z}}{ \Gamma(k+1+z)})$
$c'(z)e^x=\sum \limits_{k=-\infty}^\infty \frac{\partial}{\partial z}(\frac{x^{k+z}}{ \Gamma(k+1+z)})$
$c'(z)e^x=\sum \limits_{k=-\infty}^\infty (\frac{\ln x . x^{k+z}}{ \Gamma(k+1+z)})-\sum \limits_{k=-\infty}^\infty (\Gamma'(k+1+z)\frac{ x^{k+z}}{ \Gamma^2(k+1+z)})$
$c'(z)e^x=\ln x \sum \limits_{k=-\infty}^\infty (\frac{ x^{k+z}}{ \Gamma(k+1+z)})-\sum \limits_{k=-\infty}^\infty (\Gamma'(k+1+z)\frac{ x^{k+z}}{ \Gamma^2(k+1+z)})$
$e^x(c(z)\ln x-c'(z)) =\sum \limits_{k=-\infty}^\infty (\frac{ x^{k+z} \Gamma'(k+1+z)}{ \Gamma^2(k+1+z)})$
If we take $z=0$, we get an interesting result.
$e^x(c(0)\ln x -c'(0)) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$
$c(0)=1$ according to Equation 1
Thus
$e^x(\ln x -c'(0)) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ I do not know what $c'(0) is$.
Acoording to Norbert's answer. $c'(0) \approx -0.596347$
I have not seen that result in other place. Is it known result?Please let me know if my results are correct or not.
Can we extend all such functions that include $\Gamma(x)$ in denominator?
Thanks for advice