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)
Writing $\tan^{-1}\cot(\pi x)$ is in fact just a very fancy/silly way to write the sawtooth function.
Let $B_1(x)=s(x)=\{x\}-\frac{1}{2}$ denote the sawtooth function. Then note that $$\tan^{-1}\cot(\pi x)=\pi^2 s(x)^2.$$ Their identity is then $$\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\frac{s}{8}-\frac{s(s+1)}{2}\int_{1}^{\infty} \left(\{u\}-\frac{1}{2}\right)^2 u^{-s-2}du,$$ however, as you have noted, this isn't correct either. So what is the real identity lurking here, and how do we prove it? In fact, we have that $$\zeta(s)=\frac{s}{s-1}-\frac{1}{2}-\frac{s(s+1)}{2}\int_{1}^{\infty}\left(\{u\}^{2}-\{u\}\right)u^{-s-2}du $$ and this follows from integration by parts twice.
Writing $$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}=\int_{1}^{\infty}x^{-s}d\lfloor x\rfloor,$$ and applying integration by parts, we have $$\zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}\{u\}u^{-s-1}du.$$ Now, we could proceed from here, but instead we'll rewrite this in terms of the first periodic bernoulli polynomial $s(x)$ to obtain $$\zeta(s)=\frac{s+1}{2\left(s-1\right)}-s\int_{1}^{\infty}B_{1}(\{u\})u^{-s-1}du.$$
Applying integration by parts again, $$\int_{1}^{\infty}B_{1}(\{u\})u^{-s-1}du=u^{-s-1}\int_{1}^{u}B_{1}(\{u\})dt\biggr|_{1}^{\infty}+(s-1)\int_{1}^{\infty}u^{-s-2}\left(\int_{1}^{u}B_{1}(\{u\})dt\right)du.$$ Now, as $$\int_{0}^{x}B_{1}(\{u\})du=\frac{B_{2}\left(\{x\}\right)-B_{2}(0)}{2}=\frac{1}{2}\left\{ x\right\} ^{2}-\frac{1}{2}\left\{ x\right\},$$ we arrive at the desired identity. We could continue in this way, and rewrite the identity as $$\zeta(s)=\frac{s}{s-1}-\frac{1}{2}+\frac{s}{12}-\frac{s(s+1)}{2}\int_{1}^{\infty}B_{2}(\{u\})u^{-s-2}du,$$ applying integration by parts again, using the fact that $$\int_0^x B_n(\{t\})dt = \frac{B_{n+1}({x})-B_{n+1}(0)}{n+1},$$ and this yields a method of extending $\zeta(s)$ to the half plane $\text{Re}(s)>-n$ where the formula involves the first $n+1$ Bernoulli numbers. In fact, some combinatorial analysis can turn this into an alternate proof that $$\zeta(-n)=-\frac{B_{n+1}}{n+1},$$ and hence if combined with the functional equation for the zeta function we obtain an alternate proof for the value of $\zeta(2k)$.
Best Answer
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:
$$\int_0^{\infty} x^{s-1}e^{-nx}\,dx=\frac{1}{n^s}\int_0^\infty v^{s-1}e^{-v}dv = \frac{1}{n^s}\Gamma(s)$$