Transfer function of a classic controller
Well, that is more related to circuit theory than control. You probably know that:   


the impedance of a capacitor can be written as \$ Z_C = \frac{1}{sC}\$, 
the amp op will behave as if having a virtual ground between the two inputs,
that amp op configuration is an non-inverting amplifier.


So write down the KCL equations of the circuit, or just use the gain formula for non-inverting amplifiers (fixed in the current edit):

$$ v_i =  \frac{R_3}{R_3 + R_1 + R_2||Z_C} v_o.$$

The transfer function will be

$$ \frac{v_i}{v_o} = H^{-1}(s) =  \frac{R_3}{R_3 + R_1 + (1/R_2 +sC)^{-1}},$$

$$ H^{-1}(s) =  \frac{R}{2R + (1/R +sC)^{-1}} = \frac{R(1/R +sC)}{2R(1/R +sC) + 1}, $$

$$ H^{-1}(s) = \frac{(1 +sRC)}{2(1 +sRC) + 1} = \frac{(1 +sRC)}{(3 +s2RC)},$$

$$ H(s) = \frac{(3 +s2RC)}{(1 +sRC)} =  k_1 \frac{\frac{3}{2RC} +s}{\frac{1}{RC} +s} =  k \frac{1 + \frac{2RC}{3}s}{1 + RCs},$$

Where \$k_1\$ is some gain, and \$ k = H(0)\$. This transfer function has a pole and a zero, and it is not of the form 
$$ PID(s) = k_p + k_i/s + k_d s,$$
 since it has no purely derivative term (although in real systems they will never will due to sensor error amplification), nor it has an ideal integrator. In fact, it seems to be some compensator. 


\$ \frac{1}{CR} &gt; \frac{3}{2CR} \Leftrightarrow \$ lead compensator,
\$ \frac{1}{CR} &lt; \frac{3}{2CR} \Leftrightarrow \$ lag compensator.


The gain will be \$k = H(0) = 3\$.

Obs: 
\$R_2||Z_C\$ is the equivalent of \$R_2\$ and \$Z_C\$ in parallel.