number theory - Questions Regarding Convergence of Formulas for Riemann Zeta Function $zeta(s)$

I've been told assuming $\{x\}=x-\lfloor x\rfloor$, the following integral (1) evaluates to $\zeta(s)$ in the interval $\Re(s)\in(0,1)$.

(1) $\quad\zeta(s)=-s\int_0^\infty \{x\}\,x^{-s-1}dx\,,\quad 0<\Re(s)<1$

Since $x-\lfloor x\rfloor=SawtoothWave(x)$ and $SawtoothWave(x)=\frac{1}{2}-\frac{1}{\pi}\sum_{k=1}^\infty \frac{\sin\,(2\,\pi\,k\,x)}{k}$, this leads to the following.

(2) $\quad\zeta(s)=-s\int_0^\infty \left(\frac{1}{2}-\frac{1}{\pi}\sum_{k=1}^K \frac{\sin\,(2\,\pi\,k\,x)}{k}\right)\,x^{-s-1}dx\,,\quad 0<\Re(s)<1\ \&\ K\to\infty$

I can't seem to get integral (2) above to converge when evaluating the integral along the critical line.

Including the saw-tooth wave offset of $\frac{1}{2}$ is one problem. Integral (3) below doesn't converge, so I assume the offset is really not supposed to be included in the evaluation and am taking the approach of evaluating integral (4) below instead of integral (2) above.

(3) $\quad -s\int_0^\infty \left(\frac{1}{2}\right)\,x^{-s-1}dx\,,\quad 0<\Re(s)<1$

(4) $\quad\zeta(s)=-s\int_0^\infty \left(-\frac{1}{\pi}\sum_{k=1}^K \frac{\sin\,(2\,\pi\,k\,x)}{k}\right)\,x^{-s-1}dx\,,\quad 0<\Re(s)<1\ \&\ K\to\infty$

Question 1: Is it correct to omit the saw-tooth wave and evaluate integral (4) instead of integral (2)?

I'm using formula (5) below to evaluate integral (4) above. Increasing the evaluation limit $K$ leads to another problem. The more I increase the evaluation limit $K$, the more formula (5) seems to diverge.

(5) $\quad\zeta(s)=2^s\pi^{s-1}\,\Gamma(1-s)\sin\left(\frac{\pi\,s}{2}\right)\sum _{k=1}^K k^{s-1}\,,\quad 0

Formula (5) above is based on formula (6) below.

(6) $\quad-s\int_0^\infty \left(-\frac{\sin(2\,\pi\,k\,x)}{\pi\,k}\right)\,x^{-s-1}dx=2^s\pi^{s-1}\,\Gamma(1-s)\sin\left(\frac{\pi\,s}{2}\right)\,k^{s-1}\,,\quad 0<\Re(s)<1$

Also, note formula (5) above is consistent with the functional equation $\zeta(s)=2^s\pi^{s-1}\,\Gamma(1-s)\sin\left(\frac{\pi\,s}{2}\right)\,\zeta(1-s)$, where $\zeta(1-s)=\sum_{k=1}^\infty k^{s-1}$.

Question 2: Why does formula (5) seem to diverge as the evaluation limit $K$ is increased?

I encounter the same problem when evaluating the following integral.

(7) $\quad\zeta(s)=\frac{\zeta(s)}{\zeta(1-s)}\sum_{k=1}^K k^{s-1}\,,\quad 0

Since $\zeta(1-s)=\sum_{k=1}^\infty k^{s-1}$, it seems to me formula (7) above should converge as the evaluation limit $K$ is increased, but increasing the evaluation limit $K$ seems to have the opposite effect.

Question 3: Why does formula (7) above seem to diverge as the evaluation limit $K$ is increased?