Why is the Capacitor Inductance Positive in Transfer Function?
The transfer function of that circuit is

$$\tilde{v}_o = \frac{R}{R+1/j\omega{}C}\tilde{v}_i$$

The impedance of the capacitor is \$1/j\omega{}C\$ which is a negative imaginary number because \$1/j = -j\$.

So if you wrote the transfer function in terms of reactance, you'd have

$$\tilde{v}_o = \frac{R}{R+jX_C}\tilde{v}_i$$

because \$X_C\$ is generally taken as \$-1/\omega{}C\$.