Gauss's proof that the Digamma function equals $\int_0^{\infty}(\frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}})dt$. I was reading about the Digamma function, defined as:
$$\psi(z) = \frac{d}{dx}\ln( \Gamma(z)) = \frac{\Gamma ' (z)}{\Gamma(z)}$$
And the following integral representation of $\psi(z)$ was given for $z:\Re(z) > 0$:
$$\psi(z) = \int_0^{\infty}\Big{(}\frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}}\Big{)}dt$$
The proof of which is written off as "due to Gauss," but I cannot find his proof. What was Gauss's proof of this?
 A: You may have a look at the book Modern Analysis of Whittaker and Watson as cited in the Wikipedia article of the Digamma function. In Section 12.3 a proof is given by using the identity
$$\tag{1}\frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \frac{1}{z} + \lim_{n \rightarrow \infty} \sum_{m=1}^n \Bigg( \frac{1}{m} - \frac{1}{z+m} \Bigg)$$
together with
$$\frac{1}{z+m} = \int_0^\infty e^{-t(z+m)} \ dt.$$
In fact, this allows us to rewrite formula (1) by
$$\frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \int_0^\infty e^{-zt} \ dt + \lim_{n \rightarrow \infty} \int_0^\infty \sum_{m=1}^n (e^{-mt} - e^{-(z+m)t}) \ dt.$$
Now the sum in the integral can be simplified by using the geometric series identity. In addition, we recall that 
$$\gamma = \int_0^\infty \Bigg(\frac{1}{1-e^{-t}} - \frac{1}{t} \Bigg) e^{-t} \ dt. $$
A proof of this identity is also given in the above-mentioned book. Combining both together, we see that
\begin{align}
\frac{\Gamma'(z)}{\Gamma(z)} &= -\gamma - \lim_{n \rightarrow \infty} \int_0^\infty \frac{e^{-t}-e^{-zt}-e^{-(n+1)t}-e^{-(z+n+1)t}}{1-e^{-t}} \ dt \\
& = \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} dt - \lim_{n \rightarrow \infty} \int_0^\infty \frac{1-e^{-zt}}{1-e^{-t}} e^{-(n+1)t} \ dt.
\end{align}
By continuous continuation we know that $\frac{1-e^{-zt}}{1-e^{-t}}$ is a bounded function on $[0,1]$. Moreover, for $t \ge 1$ one has obviously that
$$ \bigg| \frac{1-e^{-zt}}{1-e^{-t}} \bigg| \leq \frac{2}{1-e^{-1}}$$
and therefore the function is bounded. Using this information, we can apply Lebesgue's theorem on dominated convergence to conclude that the integral vanishes for $n \rightarrow \infty$. All in all, we obtain that
$$\frac{\Gamma'(z)}{\Gamma(z)} = \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} dt.$$
However, I am unaware which proof was given by Gauss and it is also not mentioned in the book.
