Let $X$ and $Y$ be two random variables with density functions $f(x)$ and $g(y)$ respectively. Then distribution for random variable $Z=X+Y$ is given by convolution of $f$ and $g$. However such a computation is not defined where one of them is discrete and other one being continuous. One different approach is through their characteristic functions which are nothing but their Fourier transforms. So if $\varphi_X(t)$ and $\varphi_Y(t)$ are their respective characteristic functions (as given in https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)), then convolution can be obtained as $(f*g)(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \varphi_X(t) \varphi_Y(t) e^{-itx}dt$. I want to compute this integral somehow if $X$ is Poisson distribution (with mean=$\lambda$ and $\varphi_X(t)=e^{\lambda(e^{it}-1)}$) and $Y$ is Gamma distributed (with parameters $\theta$ and $k$ and $\varphi_Y(t) = (1-it\theta)^k$). I am unable to calculate the convolution integral. I am also unable to numerically integrate the expression because of the presence of the complex number.
Can we somehow integrate this expression (in exact sense or numerically). Or, can we use FFT and IFFT to solve the original problem since Fourier transforms are involved.
This can be solved by considering the cdf of $Z$: \begin{align*} F_Z(z)=P(Z \leq z) &= P(X+Y \leq z) \\ &= \sum_{k=0}^\infty P(X=k, X+Y \leq z)\\ &= \sum_{k=0}^\infty P(X=k, Y \leq z-k)\\ &= \sum_{k=0}^\infty p_X(k)F_Y(z-k). \end{align*} Thus, $$f_Z(z)=\frac{d}{dz}F_Z(z) = \sum_{k=0}^\infty p_X(k)f_Y(z-k).$$ This is the same answer you would get if you computed the convolution, treating $f_X(x)$ as a distribution with Dirac mass $p_X(k)$ at $x=k$ for each $k \in \mathbb{N}$.