I have the following function $$R_\infty = \sum_k P(k)[1-e^{-\lambda k\phi_\infty}$$
which is the infinite time epidemic prevalence of a network model for some degree $k$ in the network with infection rate $\lambda$, the degree distribution of the network $P(k)$, and some function $\phi_\infty$ (this is no longer relevant). In an Erdos-Renyi random network with a Poisson degree distribution, for mean degree $\mu = pN$, $P(k)=\frac{(pN)^k}{k!}e^{-pN}$. So, the expression for $R_\infty$ in the continuous case becomes
\begin{align*} R_\infty &= e^{-pN}\int \frac{(pN)^k}{k!}[1-e^{-\lambda k\phi}]dk\\\\ &= 1 - e^{-pN}\int \frac{(pN)^k}{k!}e^{-\lambda k\phi}dk \end{align*}
(since the sum of all degree probabilities is 1).
To solve this integral, I know I have to do a change of variable or something but I do not know how to proceed or how to choose what variables to change to make this a solvable integral (or at least something I could simplify for analytic interpretation). Perhaps I could let $x=\lambda k\phi$ but I don't know what else to change or how to solve this.
There is no need of integration in order to compute $R_\infty$; indeed, one has : $$ R_\infty = \sum_{k\ge0} P_\mu(k)(1-e^{-\lambda\phi k}) = 1 - e^{-\mu}\sum_{k\ge0} \frac{\mu^k}{k!}e^{-\lambda\phi k} = 1 - e^{-\mu(1-e^{-\lambda\phi})} $$