A confusion about a transfer function's Bode plot and s domain view
You see above at ω = 0 the dB is finite and in this case it is zero


No, that is not true; \$\omega\$ = 1.

Look at your graph - you are are plotting against a base of log\$_{10}(\omega)\$. 

At \$\omega\$ = 1, H(s) will be 0 dB. 0 dB = unity or 1.

For an integrator, at zero frequency, gain will be infinite because that is where the pole is.