# Show some properties of the Digamma Function

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.