I was messing around with the Gamma function and it's poles when I happened upon the following series:


Does anyone know what this is? This graph (red line) appears to be very close to the Gamma function (black line):

full graph here

enter image description here

Is there any way to turn this exactly into the Gamma function or derive the closed form of that series? I thought about adding $e^x$ to get the positive side more fixed and similar, but I'm also interested in an actual way to get this working.

  • 3
    $\begingroup$ I believe if you take what you've written subtract $f(x)$ and add $\Gamma (x)$ you will get the Gamma function as desired. But I can't be sure. I'm sorry...I couldn't help myself... $\endgroup$
    – David
    Feb 13, 2017 at 22:34

4 Answers 4


It is $f(s) = \gamma(s,1)$ the lower incomplete gamma function. You have separated $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}dx$ into $$\Gamma(s) = \int_0^1 x^{s-1} e^{-x}dx+\int_1^\infty x^{s-1} e^{-x}dx$$ where $\int_1^\infty x^{s-1} e^{-x}dx$ is entire and $\int_0^1 x^{s-1} e^{-x}dx=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{s+n-1}dx$ is a locally uniformly convergent series of poles

(using say the Riemann-Lebesgue lemma and $\Gamma(s+1) = s \Gamma(s)$ we can see that $\gamma(s,1)\to 0$ when $s$ moves away from the negative real axis)

  • $\begingroup$ Ah, that makes good sense :D $\endgroup$ Feb 13, 2017 at 22:54

Let's compute


and observe that

$$f(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n-1}[(x+n)-x]}{n!(x+n)}=1-\frac 1e+x\sum_{n=1}^\infty\frac{(-1)^n}{n!(x+n)}=1-\frac 1e+x\left(f(x)-\frac 1x\right)$$

We get :

$$\boxed{\forall x>0,\,f(x+1)=x\,f(x)-\frac 1e}$$

This is not the same functional equation than the one verified by $\Gamma$, but looks like ...

  • $\begingroup$ Oh, very interesting :D $\endgroup$ Feb 13, 2017 at 22:52
  • $\begingroup$ Beautifull answer! $\endgroup$ Aug 26, 2017 at 0:53

This series differs from $\Gamma(x)$ by an analytic function. For $\Re(x)>0$, $$ \Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \, dt \\ = \int_0^1 t^{x-1} e^{-t} \, dt + \int_1^{\infty} t^{x-1} e^{-t} \, dt \\ = f(x) + \Gamma(1,x), $$ by expanding the exponential as a power series and integrating term-by-term. $\Gamma(1,x)$ is the upper incomplete Gamma-function, and is an analytic function of $x$, while your series is the lower incomplete Gamma-function $\gamma(1,x)$. Meromorphic continuation implies that the equality derived above holds whenever $x$ is not a nonpositive integer: one can show that $f(x)$ is locally uniformly convergent on domains avoiding the nonpositive integers, so is a meromorphic function, and the result follows.

Notably, it is easy to check using integration by parts that $\Gamma(1,x)$ is exponentially small for $x \ll 0$, hence why the dominant behaviour on the left is captured by $f$. On the other hand, $f$ is much smaller than the exponentially large $\Gamma(1,x)$ for $x \gg 0$, so $\Gamma(1,x)$ is dominant on the right.

For more information, you can also see this answer I wrote a while ago.


Similar to @reuns but "from left to right".

$$\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!(n+x)}= & \int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^{n} e^{-t(n+x)}}{n!}\,dt \\ = & \int_{0}^{\infty} e^{-tx}\sum_{n=0}^{\infty}\frac{(-e^{-t})^{n} }{n!}\,dt \\ =& \int_{0}^{\infty} e^{-tx-e^{-t}}\,dt \\ & \overset{t\to -\log(s)} = \int_{0}^{1} s^{x-1} e^{-s}\,ds\\ &= \gamma(x,1) \end{align}$$

Here $\gamma(x,1)$ is the incomplete gamma function.


We can find the more general sums

$$S_{k}(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!(n+x)^{k}}, k=2, 3, ...$$

by calculating the $k-1$-st derivative of $\gamma(x,1)$


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