I have always wondered whether the Taylor Series of Gamma Function exists or not. I tried to find it, but in vain. I googled for it, but couldn't find it. Has anyone ever found its Taylor Series?

  • $\begingroup$ It certainly has Taylor series (say around $1$), but it does not have a Maclaurin series. See How to obtain the Laurent Expansion of Gamma Function Around $z=0$ for its Laurent series. $\endgroup$ Jan 24, 2018 at 1:16
  • $\begingroup$ But Laurent series has negative exponents. $\endgroup$
    – Shashank
    Jan 24, 2018 at 1:20
  • $\begingroup$ If $f(z)$ has a Laurent series with negative exponent terms around $z=a$, it cannot have a Taylor series around $z=a$. $\endgroup$ Jan 24, 2018 at 1:21
  • $\begingroup$ Really!!? Ok, thank you soo much. What about its Taylor Series around 1? What are the coefficients? $\endgroup$
    – Shashank
    Jan 24, 2018 at 1:22
  • 1
    $\begingroup$ Not a simple job either, see section 6.1 of csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf. $\endgroup$ Jan 24, 2018 at 1:45

4 Answers 4


If you want the Taylor series, you basically need the $n^{th}$ derivative of $\Gamma(x)$. These express in terms of the polygamma function. Considering

$$d_n=\frac{\left[\Gamma(x)\right]^{(n)}}{\Gamma(x)}$$ the first terms are $$d_1=\psi ^{(0)}(x)$$ $$d_2=\psi ^{(0)}(x)^2+\psi ^{(1)}(x)$$ $$d_3=\psi ^{(0)}(x)^3+3 \psi ^{(1)}(x) \psi ^{(0)}(x)+\psi ^{(2)}(x)$$ $$d_4=\psi ^{(0)}(x)^4+6 \psi ^{(1)}(x) \psi ^{(0)}(x)^2+4 \psi ^{(2)}(x) \psi ^{(0)}(x)+3 \psi ^{(1)}(x)^2+\psi ^{(3)}(x)$$ which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.

  • $\begingroup$ No, I am not looking for function evaluations as coefficients, but for the 'raw' numbers, like that in the Taylor Series of $sin(x)$. $\endgroup$
    – Shashank
    Jan 24, 2018 at 9:21
  • 3
    $\begingroup$ You simply evaluate these functions at $x=a$ $\endgroup$
    – wilsonw
    Oct 28, 2021 at 6:57

https://dlmf.nist.gov/5.7 has this:

$\dfrac1{\Gamma(z)} =\sum_{k=1}^{\infty} c_kz^k $ where $c_1=1, c_2=\gamma$, and $(k-1)c_k =\gamma c_{k-1}-\sum_{j=2}^{k-1}\zeta(j)c_{k-j} $.

It also has

$\ln(\Gamma(1+z)) =-\ln(1+z)+z(1-\gamma) +\sum_{k=2}^{\infty}(-1)^k(\zeta(k)-1)\dfrac{z^k}{k} $.


See the Euler Mascheroni Integrals $\text I_n$:

$$(-1)^n\text I_n=\int_0^\infty e^{-z}\ln^n(z)dz=\Gamma^{(n)}(1)$$


$$\Gamma(x)=\sum_{n=0}^\infty \frac{\Gamma^{(n)}(1)x^n}{n!}= \sum_{n=0}^\infty \frac{\int_0^\infty e^{-z}\ln^n(z)dz\,x^n}{n!} =\sum_{n=0}^\infty \frac{\text I_n\ (-x)^n}{n!} $$

where $(n)$ is the $n$ derivative. From the linked article:

$$\Gamma^{(0)}(1)=\text I_0=1$$

$$\Gamma^{(1)}(1)=-\text I_1=-\text{eulergamma}=-\gamma$$ $$\Gamma^{(2)}(1)=\text I_2=\gamma^2+\frac{\pi^2}6$$ $$\Gamma^{(3)}(1)=-\text I_3=-\gamma^3-\frac{\gamma\pi^2}2-2\zeta(3)$$ $$…$$

where $\zeta$ is the Riemann Zeta function,$\zeta(3)$ is Apéry’s constant, and $n\in\Bbb N$ for this problem. Please correct me and give feedback!

Here is an evaluation of the “incomplete $\text I_n$” integral with numerical proof for $n=3$,$u=4$:

$$\int_0^u e^{-t} \ln(t)^n dt=\sum_{m=0}^\infty \frac{(-1)^m}{m!}\int_0^u \ln(t)^n t^m dt=-\ln(u)^{n+1}\sum_{m=0}^\infty \frac{(-1)^m}{m!}\text E_{-n}(-(m+1)\ln(u)$$

where appears the Exponential Integral E function. See @Weijun’s comment’s CSIE link for a general integral representation of $\Gamma^{(n)}(x)$. Therefore:

$$(-1)^{n+1}\text I_n=\lim_{u\to\infty}\ln(u)^{n+1}\sum_{m=0}^\infty \frac{(-1)^m}{m!}\text E_{-n}(-(m+1)\ln(u)) $$

A series about $x=1$ is simplified as an $x!$ series. If one estimates by truncating the series and estimating $t$, then the $n,m$ upper bounds must be much greater than $t$:

$$x! = \lim_{t\to\infty} \sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^{m+1}t^{n+1}x^n\text E_{-n}(-t(m+1))}{m!n!}$$

shown here with interchangeable sums. Similarly using the CSIE link, a series about $x=a$:

$$\boxed{\Gamma(x)=\lim_{t\to\infty}\sum_{m,n=0}^\infty\frac{(-1)^{m+1}t^{n+1}\text E_{-n}(-t(m+a))(x-a)^n}{m!n!}}$$

shown here.

  • 1
    $\begingroup$ Since you're evaluating the expansion of $\Gamma(x)$ around $x=1$, the polynomial part should be $(x-1)^n$, not $x^n$. $\endgroup$ Dec 18, 2022 at 15:26
  • $\begingroup$ Or alternatively you could write that it's equal to $\Gamma(x+1)$, not $\Gamma(x)$. $\endgroup$ Dec 18, 2022 at 16:05
  • $\begingroup$ @BorisDimitrov Fixed series. Thanks for the help $\endgroup$ Dec 18, 2022 at 16:38

Let $n\ge 1$ be an integer. Then the following holds true:

\begin{equation} \lim\limits_{\xi \rightarrow 0}\frac{d^n}{d \xi^n} \left( \Gamma[\xi] - \frac{1}{\xi} \right) = \frac{1}{n+1} \cdot \sum\limits_{1\cdot m_1+2\cdot m_2+\cdots+ (n+1) \cdot m_{n+1} = n+1} \frac{(n+1)!}{\prod\limits_{j=1}^{n+1} m_j!} \cdot \prod\limits_{j=1}^{n+1} \left( \frac{\psi^{(j-1)}(1)}{j!}\right)^{m_j} \end{equation}

The Mathematica code below verifies the result for $n \le 10$.

In[3082]:= Clear[f]; xi =.; n = 9;
(*From definition*)
ex0 = Normal[Series[D[ Gamma[xi] - 1/xi, {xi, n}], {xi, 0, 0}]];

(*Using d'Hospital rule*)
ex = (D[ Exp[f[xi]], {xi, n}] + 
      xi/(n + 1) D[Exp[f[xi]], {xi, n + 1}]) /. 
    Exp[f[xi]] :> Gamma[xi] /. 
   Derivative[j_][f][xi] :> mPolyGamma[j - 1, xi];

ex = ex /. 
   mPolyGamma[m_, xi_] :> mPolyGamma[m, xi + 1] - (-1)^m m!/xi^(m + 1);

ex = Simplify[ex] /. xi Gamma[xi] :> Gamma[1 + xi] /. xi :> 0;

(*conjecture for generic n*)
ex1 = 1/(n + 1) Total[
        With[{mn = (n + 1 - Sum[eta m[eta], {eta, 1, n}])/(n + 1)}, 
            Integers], {(n + 1)!/
              m[xi]!, {xi, 1, 
               n + 1}] Product[(mPolyGamma[j - 1, 1]/j!)^m[j], {j, 1, 
               n + 1}]}, {}] /. m[n + 1] :> mn], 
         Sequence @@ 
          Table[{m[xi], 0, 
            Floor[(n + 1 - Sum[eta m[eta], {eta, 1, xi - 1}])/
             xi]}, {xi, 1, n}]]], n - 1], {}]]];
(ex - ex1) // Simplify

(ex0 - (ex /. mPolyGamma[j_, 1] :> PolyGamma[j, 1])) // Simplify

Out[3088]= 0

Out[3089]= 0

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