TITLE: $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is  it correct result?
QUESTION [7 upvotes]: $e^x=\sum \limits_{k=0}^\infty \frac{x^k}{k!}$ 
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

REPLY [4 votes]: Let 
$$\begin{eqnarray*}
g(x) &=& \sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2} \\
&=& \sum_{k=0}^\infty \frac{x^k}{k!}\psi(k+1),
\end{eqnarray*}$$
where $\psi$ is the digamma function.
Using the recurrence formula for the digamma, $\psi(x+1) = \psi(x) + 1/x$, one can show that $g(x)$ satisfies the differential equation
$$\begin{equation*}
g'(x) - g(x) = \frac{1}{x}(e^x-1).
\tag{1}
\end{equation*}$$
The boundary condition is $g(0) = \psi(1) = -\gamma$, where $\gamma$ is the  Euler-Mascheroni constant. 
(We interpret $g(0)$ as $\lim_{x\to 0^+} g(x)$.) 
The differential equation (1) can be solved using standard techniques. 
See below for a derivation.
The solution is $g(x) = e^x[\log x + E_1(x)]$,
where $E_1(x) = \int_x^\infty dt\, e^{-t}/t$ is the exponential integral.
(We assume for the moment that $|\mathrm{Arg}(x)| < \pi$.)
Thus, 
$$\begin{equation*}
\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2} 
= e^x[\log x + E_1(x)]. \tag{2}
\end{equation*}$$
This is in agreement with @Pink Elephants result.
Using the series representation for $E_1(x)$ for $|\mathrm{Arg}(x)| < \pi$
we find 
$$\begin{equation*}
\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}
= -\gamma e^x + e^x \sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k k!}. \tag{3}
\end{equation*}$$
Notice the logarithmic singularities cancel one another, so the apparent branch cut in (2) is illusory. 
The sum on the right of (3) converges for all complex $x$ and satisfies the boundary condition by inspection. 
Derivation of (2)
Using the integrating factor technique, we find the particular solution 
$$\begin{eqnarray*}
g_p(x) &=& e^{x-a} \int_{a}^x dt\, e^{-(t-a)} \frac{1}{t}(e^t-1) \\
&=& e^x[\log x + E_1(x)] - e^x[\log a + E_1(a)],
\end{eqnarray*}$$
where $|\mathrm{Arg}(a)|<\pi$. 
To satisfy the boundary condition we must add the homogeneous solution 
$$g_h(x) = e^x[\log a + E_1(a)],$$ 
so 
$$\begin{equation*}
g(x) = e^x[\log x + E_1(x)].
\end{equation*}$$
Here we use the fact that $\lim_{x\to 0^+} E_1(x) = -\log x - \gamma$, so 
$\lim_{x\to 0^+} g(x) = -\gamma$, as required.