Inverse Filter Circuit

The problem has come from the substitution of s for 2*PI*f.

s is a substitution for j*w

thus the equation actually is: 1/(1+j)
If you resolve this back to magnitude: 1+j = sqrt(1^2 + 1^2) = sqrt(2)

resulting in: 1/sqrt(2) = 0.707 as you expect

So take a low-pass filter: RC

^{simulate this circuit – Schematic created using CircuitLab}

$$ Vout = Vin*\frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} $$ $$ Vout = Vin*\frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{\sqrt{(\omega RC)^{2} + 1^{2}}} $$ From a magnitude point of view.

I would treat it like any other op-amp circuit.

Start at the right and work your way back.
(V_0-0)/R3=I0

That same current must flow from the - terminal to V1 so:
I0= (0-V1)/(R2+1/(sC2))

And the current going through R1 towards V1 is:
I1 = (Vi-V1)/R1

And the current flowing down from V1 is:
I2 = V1/(1/sC1)

Lastly, you know that the currents entering and leaving the nodes must be equal so at V1 you have:
I0+I1=I2

You should now have the equations to solve for everything in reference to Vo/Vi which is H(s)

Solving it all the way through I get this: $$ H(s)=\frac{V_o}{V_i}=\frac{-R_3C_2S}{(R_1C_1S+1)(R_2C_2S+1)+R_1C_2S} $$

Hopefully I didn't muck that up in the algebra...

From the looks of it, it looks like a bandpass filter due to the single order S term in the top.