If you let
then a simple variable substitution gives you
and thus
Now consider the geometric series
(check the signs for completeness, I don't recall all expansions in my head, the O(x^2) is also just fyi)
The above sum over the integral is convergent for $s>1$, while the expression
(check signs)
works for all complex s>1. We thus found the analytical continuation.
The integrand diverges periodically in steps of .
I'm pretty much just stepping through the start of Riemanns 1858 paper here. Knowing that exp splits into sin and cos, you can do some manipulation and end up with the famous functional equation
(The related baby version of this is Euler's reflection formula )
With this, the zeros on those negative integers are are now "trivial" since you know how Gamma and zeta on the positive side and "zeta prop sin" means these functions share their zeros.
If you want to understand how to get at the other zero's, see
https://youtu.be/Fl3XgPpvSNI 
