One wants to calculate the time dependent voltage, $V(t)$, of an electric circuit with impedance $$Z(\omega) = \frac{R + i \omega L}{(1 - \omega^2 CL) + i \omega CR}$$ and a current $$I(t) = I_0 \cos(\Omega t)$$ My initial approach was by calculating the voltage in frequency space, $V(\omega) = I(\omega)Z(\omega)$, and then carry out an inverse Fourier transform on $V(\omega)$ to get $V(t)$.
With $I(\omega) = I_0\pi\biggl(\delta(\omega - \Omega) + \delta(\omega + \Omega)\biggl)$, it gives $$V(t) = \frac{I_0}{2} \int_{-\infty}^\infty \biggl(\delta(\omega - \Omega) + \delta(\omega + \Omega)\biggl) \frac{(R + i \omega L)e^{i\omega t}}{(1 - \omega^2 CL) + i \omega CR} d\omega$$ There are 2 poles in the integrand, namely $$\omega_\pm = i\biggl(\frac{R}{2L} \pm \sqrt{\biggl(\frac{R}{2L}\biggl)^2 - \frac{1}{LC}}\biggl)$$ and therefore one can rewrite the inverse transform as $$V(t) = \frac{I_0}{2} \int_{-\infty}^\infty \biggl(\delta(\omega - \Omega) + \delta(\omega + \Omega)\biggl) \frac{(R + i \omega L)e^{i\omega t}}{(-LC)(\omega - \omega _+)(\omega - \omega _-)} d\omega$$ I would usually then solve the integral by contour integration, with a semi-circular contour closing in the positive half, where both poles will be enclosed. One can then get the answer by residue theorem.
However, what I am unsure of is how to do the contour integration when there are also delta functions involved? Since if I just solve it using the normal behaviour of delta functions, $\int_{-\infty}^\infty \delta(x-x_0)f(x) dx = f(x_0)$, it doesn't give the correct answer.
NB. I am able to get the correct answer by calculating $Z(t)$ and then get $V(t)$ by convolution theorem, but I want to be able to do it by the above method as well.