Why is the frequency domain representation of an integrator 1/s?
The integrator is part of the proportional-integral-derivative or PID block you can find in a control system. As highlighted in the above replies, the integrator "integrates" or accumulates the error over time. The error corresponds to the deviation between the target value and what the system delivers. In theory, as long as the error is present, the integrator output increases until the system corrects the deviation and cancels the static error. The below sketch shows the principle in action:



Practically speaking, an integrator can be realized around an op am with a capacitor and a resistance. The transfer function of such a block is \$H(s)=\frac{1}{s\tau}\$ in which \$\tau=R_1C_1\$ if you consider the below circuit:



The division by \$s\$ confirms the integration and represents a so-called pole at the origin. The \$\tau\$ offers a means to adjust the magnitude crossover point, the frequency at which the magnitude is 1 or 0 dB. As pointed out in some of the replies, in dc, the gain is theoretically infinite and there is no static error in the control system. Practically speaking, however, the gain is limited by the op amp open-loop gain \$A_{OL}\$ which sets a low-frequency pole as illustrated below and determined here. The magnitude in the left side exhibits a -1-slope or a 20-dB decay per decade. The phase lag in the right side is a permanent -270° or 90°. You will get more information on PIDs and their applications in switching converters here.