I came up with an integral representation for the zeta function, but haven't seen it listed online anywhere. Is it correct?
Is the following integral valid for $\Re(s)>0?$
$$\zeta(s)= \frac{1}{\Gamma(s) \Gamma(s+1)} \int_0^\infty f(t)t^{s-1}~dt$$
where,
$$f(t)=\sum_{n=1}^\infty 2\sqrt{nt}K_1(2\sqrt{nt})$$
and where,
$K_1(\cdot)$ is the modified Bessel function of the second kind.
No,
$$\zeta(s)=\frac{1}{\Gamma(s)\, \Gamma(s+1)} \int\limits_0^\infty f(t)\, t^{s-1}\,dt=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N n^{-s}\right),\quad\Re(s)>1\tag{1}$$
where
$$f(t)=\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N 2\sqrt{n t}\, K_1(2 \sqrt{n t})\right)\tag{2}$$
However, $$\zeta(s)=\frac{1}{\Gamma(s)\, \Gamma(s+1)} \int\limits_0^\infty f(t)\, t^{s-1}\,dt=\underset{N\to\infty}{\text{lim}}\left(\frac{N^{1-s}}{s-1}+\sum\limits_{n=1}^N n^{-s}\right),\quad\Re(s)>0\tag{3}$$
where
$$f(t)=\underset{N\to\infty}{\text{lim}}\left(2 N\, K_2\left(2 \sqrt{N t}\right)+\sum_{n=1}^N 2\sqrt{n t}\, K_1(2 \sqrt{n t})\right)\tag{4}$$
Also,
$$\zeta(s)=\frac{1}{\Gamma(s)\, \Gamma(s+1)} \int\limits_0^\infty f(t)\, t^{s-1}\,dt=\underset{N\to\infty}{\text{lim}}\left(\frac{N^{1-s}}{s-1}+\frac{N^{-s}}{2}+\sum\limits_{n=1}^{N-1} n^{-s}\right),\quad\Re(s)>-1\tag{5}$$
where
$$f(t)=\underset{N\to\infty}{\text{lim}}\left(2 N\, K_2\left(2 \sqrt{N t}\right)+\sqrt{N t}\, K_1\left(2 \sqrt{N t}\right)+\sum_{n=1}^{N-1} 2\sqrt{n t}\, K_1(2 \sqrt{n t})\right)\tag{6}$$