How does one arrive at the basic limit formulation for the Stieltjes constants? The starting definition that I am using is:
$$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty\gamma_n\cdot\frac{(-1)^n}{n!}(s-1)$$
If I naively differentiate, I find:
$$\zeta'(s)=\sum_{m=1}^\infty-\frac{\ln(m)}{m^s},\quad\zeta''(s)=\sum_{m=1}^\infty\frac{\ln^2(m)}{m^s},\cdots,\quad\zeta^{(n)}(s)=\sum_{m=1}^\infty(-1)^n\frac{\ln^n(m)}{m^s}$$
I say "naively", because as $s\to1^+$ (using the limit from above as this representation of $\zeta$ is not correct for $\Re s\le1$), this approach to finding the Taylor coefficients sees that:
$$\tag{1}\gamma_n=\lim_{s\to1^+}\sum_{m=1}^\infty\frac{\ln^n(m)}{m^s}$$
Except this is a divergent limit as far as I can tell. Various sources instead offer this formulation:
$$\tag{2}\gamma_n=\lim_{k\to\infty}\left[\sum_{m=1}^\infty\frac{\ln^nm}{m}-\frac{\ln^{n+1}k}{n+1}\right]$$
I simply do not see how I can go from $(1)$ to $(2)$ (or rather, how to avoid the fallacy of the divergent limit), and I have tried and failed for a while to find online proofs/resources. Note that I am not very well versed in the theory of $\zeta$; perhaps there is an analytic continuation of $\zeta$ with a more revealing form that I could use here, but I wouldn't know.
Any hints?
 A: $\newcommand{\d}{\mathrm{d}}$If as usual $\psi(x)=x-[x]-1/2$ a standard formula (see below after the answer) gives:
$$\zeta(s)=\frac{1}{s-1}-\int_{1^-}^{\infty}u^{-s}\,\d\psi(u),\,\Re s>0$$So with your notation:
$$\gamma_k=(-1)^k\frac{\d^k}{\d s^k}\left(\zeta(s)-\frac{1}{s-1}\right)\Bigg |_{s=1}=-\int_{1^-}^{\infty}u^{-1}(\log u)^k\,\d\psi(u)$$
But: $$\begin{align}\int_{1^-}^{\infty}u^{-1}(\log u)^kd\psi(u)&=\lim_{N \to \infty}\int_{1^-}^{N}u^{-1}(\log u)^k\,(\d u-\d[u])\\&=\lim_{N\to\infty}\left(\frac{(\log u)^{k+1}}{k+1}\Big |_1^N-\sum_{m=1}^N\frac {(\log m)^k}{m}\right)\end{align}$$So indeed we get the required formula:
$$\gamma_k=\lim_{N \to \infty}\left(-\frac{(\log N)^{k+1}}{k+1}+\sum_{m=1}^N\frac {(\log m)^k}{m}\right)$$
Note: since $\d\psi(u)=\d u-\d[u]$, a simple evaluation shows the formula above for $\Re s >1$ but the integral is convergent for $\Re s>0$ since $\psi$ is periodic hence $\d\psi$ has uniformly bounded integrals from $1$ to $x>1$ and then the usual Dirichlet/Abel criterion applies as long as $u^{-s} \to 0, u \to \infty$; integrating by parts, one can get the continuation of $\zeta$ to $\Re s >-1$ by a similar integral, namely $$\zeta(s)=\frac{1}{s-1}+\frac{1}{2}-s\int_{1^-}^{\infty}\psi(u)u^{-s-1}\,\d u, \Re s >-1$$
