For underlying asset $$d S = r S dt + \sigma S d W + (J-1)Sd N$$ here $W$ is a Brownian motion, $N(t)$ is Poisson process with intensity $\lambda.$
Suppose $J$ is log-normal with standard deviation $\sigma_J,$ denote $$k = E[J-1]$$ then the value of vanilla call is $$\sum\limits_{n=0}^{\infty}\dfrac{(\lambda'\tau)^n}{n!}e^{-\lambda'\tau}V_{BS}(S,t;\sigma_n,r_n)$$ Here $$\lambda' = \lambda(1+k),\ \tau =T - t$$ $$\sigma^2_n = \sigma^2 + \dfrac{n\sigma_J^2}{\tau},\ r_n=r-\lambda k+\dfrac{n\log(1+k)}{\tau}$$ and $V_{BS}$ is Black-Scholes value of a call without jumps.
The result seems to be the weighted mean of vanilla calls, but how to deduce this conclusion? Or is there any reference?
Suppose $\log J \sim N(\mu,\sigma^2_J)$ and $Z$ is standard normal, then we have \begin{eqnarray*} \log S_T &=& \log S_0 + (r - \dfrac{1}{2}\sigma^2)T + \sigma\sqrt{T} Z + N(T)\log(J)\\ &\sim& \log S_0 + \dfrac{1}{2}N(T)\sigma_J^2 +N(T)\mu + (r - \dfrac{1}{2}\hat{\sigma}^2)T + \hat{\sigma}\sqrt{T}Z\\ &=& \log \hat{S}_0 + (r - \dfrac{1}{2}\hat{\sigma}^2)T + \hat{\sigma}\sqrt{T} Z \end{eqnarray*} we omit the representations of $\hat{S}_0,\ \hat{\sigma}.$ So I think the price at $0$ should be $$\sum\limits_{n=0}^{\infty}\dfrac{(\lambda T)^n}{n!}e^{-\lambda T}V_{BS}(\hat{S}_0,0;\hat{\sigma}_n,r)$$
Why the author change the risk free rate $r_n$ and intensity $\lambda',$ does he change the measure for both Brownian motion and Poisson process?