# $e^x(\ln x-c) =\sum \limits_{k=0}^\infty \frac{ x^{k} \Gamma'(k+1)}{ (k!)^2}$ Is it correct result?

$e^x=\sum \limits_{k=0}^\infty \frac{x^k}{k!}$

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?

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Why you can extend this definition from $n\in\mathbb{N}$ to $z\in\mathbb{Z}$ ? –  Norbert Jul 21 '12 at 20:38
@Norbert You can see in my results. I derivated $\frac{d}{dx}(\sum \limits_{k=-\infty}^\infty \frac{x^{k+z}}{ \Gamma(k+1+z)})$ and I got $e^x$. –  Mathlover Jul 21 '12 at 20:42
When you prove this formula you are already assuming it is correct –  Norbert Jul 21 '12 at 20:44
@Norbert I edited my result. Thanks for your advice –  Mathlover Jul 21 '12 at 20:49
But the reciprocal of the gamma function has removable discontinuities at the negative integers, with value $0$. –  GEdgar Jul 22 '12 at 0:11

Setting aside issues of convergence for the moment, there seems to be a mistake in the formal computation. The function $\frac{\Gamma'(1+z)}{\Gamma(1+z)^2}$ extends to an entire function, whose value at $z=-n$ is $(-1)^n\Gamma(n)$. This means that, when $z=0$, we cannot ignore the terms in the sum with $k<0$. So, maybe the question should be whether $$e^x\log(x)=\sum_{k=0}^\infty \frac{x^k\Gamma'(k+1)}{(k!)^2}+\sum_{n=1}^\infty (-1)^n (n-1)! x^{-n}$$ At this point, this is all formal.

The second piece of the above series does not converge for any $x\neq 0$. However, there is a technique called Borel summation (check out W. Balser's book "From divergent power series to analytic functions", or http://en.wikipedia.org/wiki/Borel_summation), which allows one assign a "sum" to a power series whose coefficients grow factorially fast. Basically, the theory of Borel summation says that among all analytic functions in some sector having $$\sum_{n=1}^\infty (-1)^n(n-1)!x^{-n}$$ as an asymptotic expansion at infinity, there is at most one which is approximated maximally well by this expansion. This analytic function, if it exists, is called the Borel sum of the series. If we define $$g(x)=\sum_{k=0}^\infty \frac{x^k\Gamma'(k+1)}{(k!)^2},$$ and define $h(x)$ to be the Borel sum of $$\sum_{n=1}^\infty (-1)^n (n-1)! x^{-n},$$ then it makes sense to ask whether $e^x\log(x)=g(x)+h(x)$. I do not know the answer to this.

In our case, we can compute explicitly that $$h(x)=-e^x\int_x^\infty\frac{e^{-t}}{t}\,dt$$

It seems like we need to do the same thing when we write $$e^x=\sum_{k=-\infty}^{\infty}\frac{x^{k+z}}{\Gamma(k+1+z)}$$ When $z$ is not an integer, this series does not converge for any $x$ (we have that $\Gamma(k+1+z)$ decays like $\frac{1}{|k|!}$ as $k\to-\infty$). This series can be separated, the divergent part Borel summed, and again I do not know if we get the function $e^x$ in the end. I have checked that this is the case when $z$ is a half-integer (I did this computation a long time ago, and I'm only about 80% confident in its accuracy)

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Thanks a lot for your answer, Is $c(z)=1$ for $z∈C$? Do you have any suggestion to prove that? –  Mathlover Jul 21 '12 at 22:53

This is answer to previous question

This result is not correct, because for $x=1$ left hand side is $0$ while the right hand side is approximately $$\sum\limits_{k=0}^\infty \frac{\Gamma'(k+1)}{(k!)^2}\approx 0.596347$$ Problems rise even at the moment of extending $\exp$ series to negative indexes. For $k+n+1\in\mathbb{Z}_-$ the value of $\Gamma(k+n+1)$ is not well defined. So the functions $$\sum\limits_{k=0}^\infty \frac{x^{k+z}}{\Gamma(k+z+1)}\qquad\text{ and }\qquad \sum\limits_{k=0}^\infty \frac{x^{k+z}\Gamma'(k+z+1)}{\Gamma(k+z+1)^2}$$ are not defined for integer values of $z$. So you can't substitute $z=0$ in the end of your proof.

This is answer to edited question

This result still doesn't holds because limits of left hand side when $x$ tends to zero is $-\infty$ and the limit of right hand side is $$\frac{\Gamma'(0+1)}{(1!)^2}=-\gamma=-0.577216$$

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If $x=2$ then $c'(0) \approx -0.596347$ or not ? Could you please calculate it if it is easy for your calculation tool. Thanks a lot . –  Mathlover Jul 21 '12 at 22:08
I want to point out that $-e^1\int_1^\infty \frac{e^{-t}}{t}\,dt=-0.596347...$ See the expression for $h(x)$ in my answer –  Julian Rosen Jul 21 '12 at 22:25
@PinkElephants : Now, I have understood what you mean in your answer. "we cannot ignore the terms in the sum with k<0" . Yes it is my wrong suggestion. Thanks a lot for your comments and answer –  Mathlover Jul 21 '12 at 22:48
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.
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.