I'd like to know how to go about proving the result

$$\lim_{x\to-n}\dfrac{\psi(x)}{\Gamma(x)} = (-1)^{n-1}n!$$

as it appears on p.g. 2 here http://www.math.usm.edu/lambers/mat415/lecture16.pdf.

  • $\begingroup$ $\bigg(\dfrac1{\Gamma(x)}\bigg)'_{x\to-n}=(-1)^n~n!$ $\endgroup$ – Lucian Oct 15 '15 at 0:43
  • $\begingroup$ @Lucian The OP is requesting a way to prove this. $\endgroup$ – Mark Viola Oct 15 '15 at 1:09
  • $\begingroup$ @Dr.MV: The proof is trivial, and it follows from the reflection formula. $\endgroup$ – Lucian Oct 15 '15 at 1:38
  • $\begingroup$ @lucian I wouldn't say its trivial to everyone. And as far as I can see, it requires more than the reflection relation. First, one needs to extend the definition of Gamma and digamma to the left-half plane by AC. To do this, we use the functional relationship $\Gamma(x+1)=x\Gamma(x)$ repeatedly. $\endgroup$ – Mark Viola Oct 15 '15 at 2:00
  • $\begingroup$ @Dr.MV: We are asked to find $\bigg(\dfrac1{\Gamma(-x)}\bigg)'_{x\to n}$. From Euler's reflection formula, this is equivalent to evaluating $\bigg(x!~\dfrac{\sin\pi x}\pi\bigg)'_{x\to n}$. The latter has two terms, one of which vanishes. $\endgroup$ – Lucian Oct 15 '15 at 2:08

Recall that the analytic continuation of the Gamma function can be obtained by using repeatedly the functional relationship $\Gamma(x+1)=x\Gamma(x)$. Then, for $x+m+1>0$, we can write

$$\Gamma (x)=\frac{\Gamma(x+m+1)}{x(x+1)(x+2)\cdots(x+m-1)(x+m)} \tag 1$$

Using $(1)$ reveals that for $x+m+1>0$, the digamma function can be expressed as

$$\psi(x)=\frac{d\log \Gamma (x)}{dx}=\psi(x+m+1)-\sum_{k=0}^m\frac{1}{x+k} \tag 2$$

Therefore, using $(1)$ and $(2)$, the ratio $\frac{\psi(x)}{\Gamma(x)}$ is given by

$$\frac{\psi(x)}{\Gamma(x)}=\left(\psi(x+m+1)-\sum_{k=0}^m\frac{1}{x+k}\right)\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\right) \tag 3$$

For $0<n\le m$, we note that for $x=-n$, $\Gamma(x+m+1)=(m-n)!$ and $\psi(x+m+1)=H_{m+1-n}-\gamma$. Therefore, we have

$$\begin{align} \lim_{x\to -n}\frac{\psi(x)}{\Gamma(x)}&=\lim_{x\to -n}\left(\frac{\psi(x+m+1)}{\Gamma(x+m+1)}\left(x(x+1)(x+2)\cdots(x+m-1)(x+m)\right)\right)\\\\ &-\lim_{x\to -n}\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\sum_{k=0}^{m}\frac{1}{x+k}\right)\\\\ &=-\lim_{x\to -n}\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\sum_{k=0}^{m}\frac{1}{x+k}\right)\\\\ &=-\frac{1}{(m-n)!}\lim_{x\to -n}\sum_{k=0}^m\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{x+k}\\\\ &=-\frac{1}{(m-n)!}\left((-n)(-n+1)(-n+2)\cdots (-2)(-1)(1)(2)\cdots(m-n-1)(m-n)\right)\\\\ &=-(1)^{n+1}n! \end{align}$$

as was to be shown!

  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Oct 15 '15 at 14:41
  • $\begingroup$ Thank you very much -- this is exactly the sort of thing I was looking for. From the main body of working could you please tell me how the first of the limit terms is eliminated and how the penultimate line follows from the previous. $\endgroup$ – bfdes Oct 15 '15 at 23:26
  • $\begingroup$ You're welcome. My pleasure. Since $x\to -n$ all terms without a $1/(x+n)$ go to zero. And similarly, we arrive at the pentultimate expression follow since the $(x+n)$ term in the numerator is precisely compensated by the term for $k=n$ in the summation. $\endgroup$ – Mark Viola Oct 16 '15 at 1:14
  • $\begingroup$ Understood, thank you. I'm not familiar with many of the other tools Lucien mentioned to prove the result more trivially. $\endgroup$ – bfdes Oct 16 '15 at 9:13
  • $\begingroup$ Yes. I wanted to present a way forward that did not rely on a relationship (i.e., the reflection formula) that is not easy to intuitively understand or prove with standard tools. $\endgroup$ – Mark Viola Oct 16 '15 at 14:26

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