To evaluate that limit, we can expand each function in a Laurent series at $s=0$ and use the following 3 facts about the Hurwitz zeta function:
$$ \zeta(-s,a) = \zeta(-s,a+1) + a^{s} \tag{1}$$
$$ \zeta'(-s,a) = \zeta'(-s,a+1) -a^{s} \log(a) $$
$$\zeta(-n, a) = -\frac{B_{n+1}(a)}{n+1} \ , \ n \in\mathbb{N} \tag{2}$$
Doing so, we get
$$z - z \log z - \frac{\gamma z^{2}}{2} + \lim_{s \to 0^{+}} \Big[ - \Gamma(s-1) \zeta(s-1,z+1) -z \Gamma(s) \zeta(s) + \frac{z^{2}}{2} \Gamma(s+1) \zeta(s+1)$$
$$+ \Gamma(s-1) \zeta(s-1) \Big]$$
$$ = z - z \log z - \frac{\gamma z^{2}}{2}$$
$$ + \lim_{s \to 0^{+}} \Bigg[-\Big(-\frac{1}{s} + \gamma -1 + \mathcal{O}(s) \Big) \Big( -\frac{z^{2}}{2}+\frac{z}{2}-\frac{1}{12}-z + \zeta'(-1,z)s + z \log z \ s + \mathcal{O}(s^{2}) \Big)$$
$$ - z \Big( \frac{1}{s} - \gamma + \mathcal{O}(s) \Big) \Big( - \frac{1}{2} - \frac{\log (2 \pi)}{2} s + \mathcal{O}(s^{2}) \Big) + \frac{z^{2}}{2} \Big(1- \gamma s + \mathcal{O}(s^{2}) \Big) \Big( \frac{1}{s} + \gamma + \mathcal{O} (s) \Big) $$
$$ + \Big(- \frac{1}{s} + \gamma -1 + \mathcal{O} (s) \Big) \Big( - \frac{1}{12} + \zeta'(-1) s + \mathcal{O}(s^{2}) \Big) \Bigg] $$
$$ = z - z \log z - \frac{\gamma z^{2}}{2}$$
$$ + \lim_{s \to 0^{+}} \Big[\zeta'(-1,z) + z \log z + \frac{\gamma z^{2}}{2} - \frac{z^{2}}{2} + \frac{z}{2} - z + \frac{z \log(2 \pi)}{2} - \zeta(-1)+ \mathcal{O}(s) \Big] $$
$$ = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} -\zeta'(-1)+ \zeta'(-1,z) $$
$ $
$(1)$ http://dlmf.nist.gov/25.11 (25.11.3)
$(2)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (9)
Recall that for $t>1$, $$\frac{1}{t-1}=\sum_{n=1}^\infty t^{-n}$$
Then substitute $t=e^{x}$ for $x>0$.
Then substituting $x=\frac{v}{n}$ in the $n$th term of the integral, you get:
$$\int_0^{\infty} x^{s-1}e^{-nx}\,dx=\frac{1}{n^s}\int_0^\infty v^{s-1}e^{-v}dv = \frac{1}{n^s}\Gamma(s)$$
Best Answer
The correct way to derive the formula is \begin{align*} \Gamma (s)\zeta (s) = \sum\limits_{n = 1}^\infty {\frac{{\Gamma (s)}}{{n^s }}} = \sum\limits_{n = 1}^\infty {\int_0^{ + \infty } {\mathrm{e}^{ - t} \frac{{t^{s - 1} }}{{n^s }}\mathrm{d}t} }&\, \mathop = \limits^{t = nx} \,\sum\limits_{n = 1}^\infty {\int_0^{ + \infty } {\mathrm{e}^{ - nx} x^{s - 1}\, \mathrm{d}x} } \\ &\;\, = \int_0^{ + \infty } {\sum\limits_{n = 1}^\infty {\mathrm{e}^{ - nx} } x^{s - 1} \,\mathrm{d}x} . \end{align*}