Deriving Bandpass Transfer Function
The TF of the circuit is -\$\dfrac{Z_f}{Z_i}\$ where \$Z_f\$ = \$R_2\$||X\$_{C_2}\$ and \$Z_i\$ = \$R_1\$ +X\$_{C_1}\$.

I.e. \$Z_f\$ is the feedback impedance and \$Z_i\$ is the input impedance. 

In terms of the s-plane operator: -

\$Z_f\$ = \$\dfrac{R_2\cdot \frac{1}{sC_2}}{R_2 + \frac{1}{sC_2}}\$ and \$Z_i\$ = \$R_1+\frac{1}{sC_1}\$

The TF then becomes \$-\dfrac{\dfrac{R_2\cdot \frac{1}{sC_2}}{R_2 + \frac{1}{sC_2}}}{R_1+\frac{1}{sC_1}}\$

If you work this down you get TF = \$\dfrac{-R_2}{1+sC_2R_2}\cdot\dfrac{sC_1}{1+sC_1R_1}\$

I think you can see that this pretty much aligns with the TF at the top of the question (when s is replaced by jw where w = 2\$\pi f\$)