Digamma Identities
Non-detailed proof: (copied from my old answer here. Thinking of adding more detail and thus copying this if its not clear enough)
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 $$ \psi(z)= \int_{0}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du. $$
Let $F(x)$ be represented by the integral
$$F(x)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}\right)\,dt\tag 1$$
Differentiating under the integral in $(1)$, we find that
$$\begin{align} F'(x)&=\int_0^\infty \frac{te^{-xt}}{1-e^{-t}}\,dt\\\\ &=\sum_{n=0}^\infty \int_0^\infty te^{-(n+x)t}\,dt\\\\ &=\sum_{n=0}^\infty \frac{1}{(n+x)^2}\tag 2 \end{align}$$
Next, integrating both sides of $(2)$ reveals
$$\begin{align} F(x)-F(1)&=\int_1^x F'(t)\,dt\\\\ &=\sum_{n=0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+x}\right)\\\\ F(x)&=F(1)+\sum_{n=0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+x}\right)\tag3 \end{align}$$
where $F(1)=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}\right)\,dt$.
Then, we can evaluate $F(1)$ by simple application of integration by parts. Proceeding, we find that
$$\begin{align} F(1)&=\int_0^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}\right)\,dt\\\\ &=\lim_{\epsilon\to 0^+}\left(\int_\epsilon^\infty \left(\frac{e^{-t}}{t}-\frac{e^{-t}}{1-e^{-t}}\right)\,dt\right)\\\\ &=\lim_{\epsilon\to 0^+}\left(-\log(\epsilon)+\int_\epsilon^\infty \log(t)e^{-t}\,dt+\log(1-e^{-\epsilon})\right)\\\\ &=\int_0^\infty \log(t)e^{-t}\,dt\tag4\\\\ &=-\gamma \tag5 \end{align}$$
where in going from $(4)$ to $(5)$, I appealed to the note at the end of THIS ANSWER.
Finally, substituting $(5)$ in $(3)$ yields
$$F(x)=-\gamma +\sum_{n=0}^\infty\left( \frac{1}{n+1}-\frac{1}{n+x}\right)=\psi(x)$$
as was to be shown!