To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$ N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi $$ where $f(\phi)$ is the characteristic function of the standard normal distribution: $$ f(\phi) = e^{-\frac{1}{2}\sigma^2}. $$ Hence, the Black-Scholes formula becomes $$ C(S,t) = S P_1 - K e^{-r(\tau)} P_2, $$ where $$ P_j = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi d_j} f(\phi)}{i\phi}] d\phi \\ f_j = e^{-\frac{1}{2}\phi^2}, \\ d_1 = \frac{\ln(\frac{S}{K})+(r+\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} \\ d_2 = \frac{\ln(\frac{S}{K})+(r-\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} = d_{1}-\sqrt{\nu_0}\sqrt{\tau}. $$

The distribution in the Black-Scholes model is standardized whereas in the Heston model, it is not. It may be easier to first transform the $P_j$ terms so that they have a similar structure. With $x=\log(S)$, the $P_j$ terms can be rewritten as: $$ P_1 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_1(\phi)}{i\phi}] d\phi \\ f_1 = e^{\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}, \\ P_2 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_2(\phi)}{i\phi}] d\phi \\ f_2 = e^{-\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}. \\ $$ In this form, you should be able to show that the Heston model converges to the Black-Scholes model as $\sigma \to 0$ when $\theta = v_0$ (although the details may be rather tedious to go through...). You may want to try equivalent formulations of the solution, other than Heston's original (e.g. the form in Gatheral: The Volatility Surface or Cont & Tankov: Financial modeling with Jump Processes), which avoid the division by zero problem a bit.

To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$ N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi $$ where $f(\phi)$ is the characteristic function of the standard normal distribution: $$ f(\phi) = e^{-\frac{1}{2}\phi^2}. $$ Hence, the Black-Scholes formula becomes $$ C(S,t) = S P_1 - K e^{-r(\tau)} P_2, $$ where $$ P_j = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi d_j} f(\phi)}{i\phi}] d\phi \\ f_j = e^{-\frac{1}{2}\phi^2}, \\ d_1 = \frac{\ln(\frac{S}{K})+(r+\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} \\ d_2 = \frac{\ln(\frac{S}{K})+(r-\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} = d_{1}-\sqrt{\nu_0}\sqrt{\tau}. $$

The distribution in the Black-Scholes model is standardized whereas in the Heston model, it is not. It may be easier to first transform the $P_j$ terms so that they have a similar structure. With $x=\log(S)$, the $P_j$ terms can be rewritten as: $$ P_1 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_1(\phi)}{i\phi}] d\phi \\ f_1 = e^{\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}, \\ P_2 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_2(\phi)}{i\phi}] d\phi \\ f_2 = e^{-\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}. \\ $$ In this form, you should be able to show that the Heston model converges to the Black-Scholes model as $\sigma \to 0$ when $\theta = v_0$ (although the details may be rather tedious to go through...). You may want to try equivalent formulations of the solution, other than Heston's original (e.g. the form in Gatheral: The Volatility Surface or Cont & Tankov: Financial modeling with Jump Processes), which avoid the division by zero problem a bit.

To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$ N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi $$ where $f(\phi)$ is the characteristic function of the standard normal distribution: $$ f(\phi) = e^{-\frac{1}{2}\sigma^2}. $$ Hence, the Black-Scholes formula becomes $$ C(S,t) = S P_1 - K e^{-r(\tau)} P_2, $$ where $$ P_j = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi d_j} f(\phi)}{i\phi}] d\phi f_j = e^{-\frac{1}{2}\phi^2}, \\ d_1 = \frac{\ln(\frac{S}{K})+(r+\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} \\ d_2 = \frac{\ln(\frac{S}{K})+(r-\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} = d_{1}-\sqrt{\nu_0}\sqrt{\tau}. $$

The distribution in the Black-Scholes model is standardized whereas in the Heston model, it is not. It may be easier to first transform the $P_j$ terms so that they have a similar structure. With $x=\log(S)$, the $P_j$ terms can be rewritten as: $$ P_1 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_1(\phi)}{i\phi}] d\phi \\ f_1 = e^{\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}, \\ P_2 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_2(\phi)}{i\phi}] d\phi \\ f_2 = e^{-\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}. \\ $$ In this form, you should be able to show that the Heston model converges to the Black-Scholes model as $\sigma \to 0$ when $\theta = v_0$ (although the details may be rather tedious to go through...). You may want to try equivalent formulations of the solution, other than Heston's original (e.g. the form in Gatheral: The Volatility Surface or Cont & Tankov: Financial modeling with Jump Processes), which avoid the division by zero problem a bit.

To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$ N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi $$ where $f(\phi)$ is the characteristic function of the standard normal distribution: $$ f(\phi) = e^{-\frac{1}{2}\sigma^2}. $$ Hence, the Black-Scholes formula becomes $$ C(S,t) = S P_1 - K e^{-r(\tau)} P_2, $$ where $$ P_j = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi d_j} f(\phi)}{i\phi}] d\phi \\ f_j = e^{-\frac{1}{2}\phi^2}, \\ d_1 = \frac{\ln(\frac{S}{K})+(r+\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} \\ d_2 = \frac{\ln(\frac{S}{K})+(r-\frac{\nu_0}{2})(\tau)}{\sqrt{\nu_0}\sqrt{\tau}} = d_{1}-\sqrt{\nu_0}\sqrt{\tau}. $$

The distribution in the Black-Scholes model is standardized whereas in the Heston model, it is not. It may be easier to first transform the $P_j$ terms so that they have a similar structure. With $x=\log(S)$, the $P_j$ terms can be rewritten as: $$ P_1 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_1(\phi)}{i\phi}] d\phi \\ f_1 = e^{\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}, \\ P_2 = \frac{1}{2} + \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi \log(K)} e^{i\phi x + i\phi r \tau } f_2(\phi)}{i\phi}] d\phi \\ f_2 = e^{-\frac{1}{2} i v_0\tau\phi -\frac{1}{2} v_0 \tau \phi^2}. \\ $$ In this form, you should be able to show that the Heston model converges to the Black-Scholes model as $\sigma \to 0$ when $\theta = v_0$ (although the details may be rather tedious to go through...). You may want to try equivalent formulations of the solution, other than Heston's original (e.g. the form in Gatheral: The Volatility Surface or Cont & Tankov: Financial modeling with Jump Processes), which avoid the division by zero problem a bit.