Given formula, for $s \in \mathbb{N}$, $s \geq 2$ should be equal to;
$$ \zeta(s)= \frac{(-2 \pi)^{s-1}}{\Gamma(s)} i\int_{0}^1(1-u)^{-1} u^{s-1} \left[e^{\left(2 \pi u+\frac{\pi s}{2}\right)}Li_{1-s}\left(e^{2\pi i u}\right)-e^{-i\left(2 \pi u+\frac{\pi s}{2}\right)}Li_{1-s}\left(e^{-2 \pi i u} \right)\right]du$$
Here is the method and all explanations;
Deleting Gamma's function singularity
I will change derivative notation into repeated integration notation, which will be showed as indefinite integral, which is just notation convention. Because derivatives are local operators, it will do not change result's, if that substitution will be made;
$$\displaystyle \lim\limits_{l \rightarrow 1}\lim\limits_{t \rightarrow 1} \Gamma(1-s)\frac{d^{s-1}}{dt^{s-1}} \left( \cos(2\pi t l)-\sin(2 \pi t l ) \cot(\pi t) \right)= \lim\limits_{l \rightarrow 1}\lim\limits_{t \rightarrow 1} \Gamma(1-s) \underbrace{\int \int ...\int}_{1-s} \left( \cos(2\pi t l)-\sin(2 \pi t l ) \cot(\pi t) \right) \underbrace{ dt \ dt \ ... \ dt}_{1-s}$$
We can use Cauchy's equation for repeated integration and shift domain of convergence of derived equation by deriving it with respect to t. Let's keep in mind that derivatives are local operator, so lower boundary of integration does not matter. Let's derive firstly general formula and than apply it to our equation;
$$ \displaystyle \leftindex_{0}{I}_{t}^{x} f(t)= \frac{1}{\Gamma(x)}\int_{0}^t(t-u)^{x-1}f(u)du=\frac{1}{\Gamma(x)}\int_{0}^{1}(1-u)^{x-1}t^xf(tu)du$$
Formula written in such way, allows us to derivative both sides with respect to $t$, which shifts degree of fractional calculus by one. However we can generalise it by deriving equation with degree $s$, which for $s-x \in \mathbb{N}$, by the power of locality of derivative of fractional calculus, will make the equation equal to standard derivative of degree $s-x$.
$$ \displaystyle \leftindex_{0}{I}_{t}^{x-s} f (t)=\frac{d^{s-x}}{dt^{s-x} }f(t)= \frac{1}{\Gamma(x)}\int_{0}^{1} \frac{d^s}{dt^s}(1-u)^{x-1}t^xf(tu)du, \ s-x \in \mathbb{N}$$
Let's use Leibniz general rule for derivative of product of two functions $\frac{d^s}{dt^s}f(t)g(t)=\sum_{k=0}^{\infty} {{s}\choose{k}}\frac{d^{s-k}}{dt^{s-k}}f(t)\frac{d^k}{dt^k}g(t)$. However, let's keep by now solution for all $s$, until we do not use limit for $s$ going to some natural number;
$$\displaystyle \frac{d^{s-x}}{dt^{s-x} }f(t)=\frac{1}{\Gamma(x)} \int_{0}^{1}(1-u)^{x-1}\sum_{k=0}^{s}{{s}\choose{k}}\frac{t^{x-s+k}x!}{(x-s+k)!} f^{(k)}\left(tu\right)u^k du, \ s-x \in \mathbb{N}$$
We can set $t=1$ and write limit of $x \rightarrow 0$. However I will change $f^{(n)}(u)$ into $\frac{d}{du} f(u)$, because of reasons to have possibility to put to equation function we need and derivative it. Final form of the tools we need, represents as;
$$\displaystyle \lim\limits_{t \rightarrow 1} \frac{d^{s}}{dt^{s} }f(t)=\lim\limits_{x \rightarrow 0} \frac{1}{\Gamma(x)} \int_{0}^{1}(1-u)^{x-1}\sum_{k=0}^{s}{{s}\choose{k}}\frac{x!}{(x-s+k)!} u^k \frac{d^k}{du^k}f\left(u\right) du$$
In such case for $s$ going to some natural number, where equation holds, we can write down;
$$\displaystyle \zeta(s)= \lim\limits_{l \rightarrow 1} \lim\limits_{(x,s) \rightarrow {0,\mathbb{N}}} \frac{\Gamma(1-s)}{\Gamma(x)} \int_{0}^1(1-u)^{x-1}\sum_{k=0}^{s-1}{{s-1}\choose{k}}\frac{x!}{(x-s-1+k)!} u^k \frac{d^k}{du^k} \left( \cos(2\pi u l)-\sin(2 \pi u l ) \cot(\pi u) \right) du $$
We can notice that, if we let $x$ to go to 0, as $s$ goes to natural number, singularities of gammas functions will delay and we will be left with formula written down below;
$$\zeta(s)= \lim\limits_{l \rightarrow 1} \frac{(-1)^{s-1}}{\Gamma(s)} \int_{0}^1(1-u)^{-1} u^{s-1} \frac{d^{s-1}}{du^{s-1}} \left( \cos(2\pi u l)-\sin(2 \pi u l ) \cot(\pi u) \right) du $$
Limit of derivative of trigonometric component
By now, we can use Leibniz general rule again to obtain derivative of given if trigonometric formula;
$$\displaystyle \lim\limits_{l \rightarrow 1} \frac{d^{s-1}}{du^{s-1}} \left( \cos(2\pi u l)-\sin(2 \pi u l ) \cot(\pi u) \right)= \lim\limits_{l \rightarrow 1} \left[\frac{d^{s-1}}{du^{s-1}}\cos \left(2 \pi l u\right) -\sum_{k=0}^{\infty} {{s-1}\choose{k}}\frac{du^{s-1-k}}{du^{s-1-k}}\sin(2 \pi l u) \frac{du^k}{du^k} \cot(\pi u)\right]$$
We can replace reciprocal of gamma function to inverse Laplace transform of power function by using relation $\int_{\alpha-i\infty}^{\alpha + i \infty} e^{tx}x^s=\frac{x^{-1-s}}{(-1-s)!}$. However, integral will be used symbolically, as some transform, so there will be no need to check ratio of convergence. Moreover it will be used Euler's identity to group elements, and prepare it to obtain result from summation. In such case we obtain;
$$ (2\pi )^{s-1}\left[\sin \left(2 \pi u+\frac{\pi s}{2}\right) -\Gamma(s)\int_{\alpha-i\infty}^{\alpha + i \infty} e^{t}t^{-s} \frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{k!}\left(e^{i\left(2 \pi u+\frac{\pi s}{2}\right)}\frac{-it}{2}^k+e^{-i\left(2 \pi u+\frac{\pi s}{2}\right)}\frac{it}{2}^k\right) \cot^{(k)}(\pi u)dt\right]$$
That will allow us to notice, Taylor series in formula, which relates function $f(x)$, to it's derivatives. That observation allows us to write down;
$$ \displaystyle (2\pi )^{s-1}\left[\sin \left(2 \pi u+\frac{\pi s}{2}\right) -\Gamma(s)\int_{\alpha-i\infty}^{\alpha + i \infty} e^{t}t^{-s} \frac{1}{2}\left(e^{i\left(2 \pi u+\frac{\pi s}{2}\right)}\cot\left(\pi u-\frac{it}{2}\right)-e^{-i\left(2 \pi u+\frac{\pi s}{2}\right)}\cot\left(-\pi u-\frac{it}{2}\right)dt\right)\right]$$
We can use Euler identity again, to rewrite $\cot(x)=\frac{i\left(e^{ix}+e^{-ix} \right)}{e^{ix}-e^{-ix}}=i\left(1-2\sum_{n=0}^{\infty}e^{2x i}\right)$. That form of equation will allow us to solve, symbolically given inverse Laplace transform. In such case we can write that;
$$\displaystyle (2\pi )^{s-1}\left[\sin \left(2 \pi u+\frac{\pi s}{2}\right) -\Gamma(s)\int_{\alpha-i\infty}^{\alpha + i \infty} e^{t}t^{-s} \frac{i}{2}\left(e^{i\left(2 \pi u+\frac{\pi s}{2}\right)}\left(1-2\sum_{n=0}^{\infty}e^{2ni\left(\pi u-\frac{it}{2}\right)}\right)-e^{-i\left(2 \pi u+\frac{\pi s}{2}\right)}\left(1-2\sum_{n=0}^{\infty}e^{2ni\left(-\pi u-\frac{it}{2}\right)}\right)dt\right)\right]$$
By now we can recognise here, inverse Laplace transform for power function. In such case we can write down;
$$\displaystyle (2\pi )^{s-1}\left[\sin \left(2 \pi u+\frac{\pi s}{2}\right) -\frac{i}{2}\left(e^{i\left(2 \pi u+\frac{\pi s}{2}\right)}\left(1-2\sum_{n=0}^{\infty}e^{2\pi i n u}(n+1)^{s-1}\right)-e^{-i\left(2 \pi u+\frac{\pi s}{2}\right)}\left(1-2\sum_{n=0}^{\infty}e^{-2 \pi i n u} (n+1)^{s-1}\right)\right)\right]$$
We can recognise written series as Poly-logarithm, where $Li_{s}(x)=\sum_{n=1}^{\infty}\frac{x^n}{n^s}$, which gives us final result of limit of derivative of trigonometric component of given integral;
$$ \displaystyle (2\pi )^{s-1}i\left[e^{i\left(2 \pi u+\frac{\pi s}{2}\right)}Li_{1-s}\left(e^{2\pi i u}\right)-e^{-i\left(2 \pi u+\frac{\pi s}{2}\right)}Li_{1-s}\left(e^{-2 \pi i u} \right)\right]$$
