Let $\psi(z)$ denote the Digamma function, $\psi(z)=\frac{d}{dz}\ln \Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}$. I am meant to show the following properties of $\psi$:

  1. $\psi$ is meromorphic in $\mathbb{C}\backslash\{0,-1,-2,-3,...\}$
  2. $\psi(1)$=$-\gamma$ where $\gamma$ is the euler mascheroni constant.
  3. $\psi(z)=-\gamma-\frac{1}{z}-\sum\limits_{\nu=1}^\infty \left(\frac{1}{z+\nu}-\frac{1}{\nu}\right)$

So far I am quite stuck on what to do. I know about the completion formula, stirling's formula, and the duplication formula but none of those appear to apply. I cannot find a way to characterize $\Gamma'(z)$ either. Any tips, hints, or suggestions would be much appreciated!



$1$. Prove that $\Gamma(z)$ has no zeroes so $\psi(z)$ can have no singular points other than the poles $0,-1,-2,...$. Then it's not too hard to prove using the properties of $\Gamma$ function that $\psi$ has the representation $$ \psi(z)=-\frac{1}{z+n}+\mathcal{E}(z+n) $$ Here $\mathcal{E}(z+n)$ is the regular part of $\psi(z)$.

$2$. See the comment.

$3.$ Non-detailed proof:

Using the definition $\Gamma'(z)$ and by replacing the $\text{log}(t)$ term with the integral $$ \text{log}(t)= \int_0^{\infty}\frac{e^{-x}-e^{-xt}}{x}, dx $$ we can show that $$ \Gamma'(z)=\int_0^\infty \frac{dx}{x}\left[ e^{-x}\Gamma(z)-\int_0^\infty e^{-t(x+1)}t^{z-1} dt \right]. $$ Using the substitution $u=t(x+1)$ we get $$ \psi(z)=\int_0^\infty\left[e^{-x}-\frac{1}{(x+1)^{z}} \right]\frac{dx}{x}. $$ Using this we can find $$ \psi(z)=\lim_{\epsilon \to 0} \left[ \int_\epsilon^\infty \frac{e^{-x}}{x}dx-\int_\epsilon^\infty \frac{1}{(x+1)^{z}x} dx \right]. $$ Substituting $1+x = e^u$ for the last integral we get $$ \psi(z)=\lim_{\epsilon \to 0} \left[ \int_{log(1+\epsilon)}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du -\int_{log(1+\epsilon)}^\epsilon \frac{e^{-u}}{u} du \right]. $$ Here the last integral goes to $0$ as $\epsilon \to 0$ so we get $$ \psi(z)= \int_{0}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du. $$ Plugging in $z=1$, subtracting the result and using substitution $x=e^{-u}$ we finally get $$ \psi(z)=-\gamma+\int_0^{1}\frac{1-x^{z-1}}{1-x}dx. $$ Now the claim follows by using $(1-x)^{-1}=\sum_{n=0}^{\infty} x^n$ and integrating term by term.

  • $\begingroup$ This was hard to follow but it definitely works, thanks. $\endgroup$ – TSF Mar 2 '16 at 16:51
  • 1
    $\begingroup$ @TonyS.F. Well I didnt add the details on purpose so you can try to fill in the blank spots and yeah there is a lot to fill. :) $\endgroup$ – Harto Saarinen Mar 2 '16 at 17:38

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