I was trying to perform the contour integral of the digamma function $\oint\limits_C \psi(z)\,dz$ on the neighborhood (a small circle $-k+re^{it}$, $k \in \mathbb{Z}$ ) of $k$, before actually realizing that due to the residue theorem $\operatorname{res}(\psi(z),-k)=\frac{1}{2\pi i}\oint\limits_C \psi(z)\,dz=-1$.

Now I know the answer, nevertheless I'm still curious as how this could be done by directly integrating.

I know that $\int \psi(z)\,dz=\log\Gamma(z)$, so $$\int_{0}^{2\pi} \psi(-k+re^{it})ire^{it}\,dt=\log\Gamma(\frac{k+re^{2 \pi}}{k+re^{0}})$$ but integrating between $0$ and $2\pi$ would just give zero as result (due to the symmetry int he function?) so I divided the integration limits: $$2\int_{0}^{\pi} \psi(-k+re^{it})ire^{it}\,dt=2\log\Gamma(\frac{k+re^{ \pi}}{k+re^{0}})$$ When I do numerical approximations to this I do get the result I'm looking for, i.e. $-2\pi i$, but I do not know how to formalize this calculation on $\lim_{r\rightarrow0}$. Could someone please help me?

Thanks in advance for your ideas!


You need to be more meticulous if you attempt to use fundamental theorem of calculus for contour integrals.

As an example, because you know $\int\frac1zdz=\ln z$, you can also similarly show $\oint\frac1zdz=0$ over an unit square contour.

Note that evaluating $[f(c+re^{it}]^{2\pi}_{t=0}$ is equivalent to $f(z)$ ‘going around’ $c$ once.

The problem is $\ln\Gamma(z)$ has a branch point at $z=-k$. No matter how you choose the branch cut, going around the branch point is non zero.

Here, I will take $\ln z=\ln|z|+i\arg z$ with $\arg\in[0,2\pi)$.

We have $$\int_{|z+k|=r}\phi(z)dz=[\ln\Gamma(-k+re^{it})]^{2\pi}_0$$

Firstly, we want to make our function easier for analysis: $$\ln\Gamma(-k+re^{it})=\ln\Gamma(1+re^{it})-\ln(re^{it})-\sum^k_{n=1}\ln(re^{it}-n)$$ due to the formula $\ln\Gamma(z)=\ln\Gamma(z+1)-\ln z$.

Except the second term, each term ‘going around’ once would give zero.

(e.g. The first term is $\ln\Gamma(z)$ going around $1$ once; since $\ln\Gamma(z)$ is holomorphic around there, going around once gives $0$.)

Then, clearly, $$[\ln\Gamma(-k+re^{it})]^{2\pi}_0=2\pi i$$

*It might be somewhat unclear to use the term ‘going around’. Please let me know if you don’t understand something.

  • $\begingroup$ Great, thanks I'm still on an early learing stage of analysis, and this was very helpful. $\endgroup$ – firulander Aug 12 '18 at 14:18
  • $\begingroup$ @firulander You may also want to take a look at my very first question on Math SE: math.stackexchange.com/questions/2559475/… $\endgroup$ – Szeto Aug 12 '18 at 14:56
  • $\begingroup$ @firulander Everyone starts from zero; do not give up and have fun in complex analysis! $\endgroup$ – Szeto Aug 12 '18 at 14:57

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