In Ronkin's Теория аналитических функций нескольких переменных he introduces an integral $$f(z)=-\frac1\pi\int_0^\infty\int_0^{2\pi}\varphi\left(z+t\mathrm{e}^{-i\theta}\right)\mathrm d\theta\mathrm dt$$ where $z\in\mathbb{C}$, $\varphi\in C^k(\mathbb{R}^2)$ is a complex function that is $C^k$ as a function of two real variables, and $\mathrm{supp}\,\varphi$ is compact. He says the function $\varphi(z+t\mathrm e^{i\theta})$ treated as a function of parameter $z$ is $C^k$ (which again is to be understood as a function of two real variables). He then says the integral, or the function $f$ is also $C^k(\mathbb{R}^2)$ because the support of $\varphi$ is compact.
I don't understand the latter statement, why is it $C^k(\mathbb{R}^2)$? Where do we need the compactness of $\mathrm{supp}\,\varphi$?
To verify the smoothness, you just need to ascertain that it is "ok to differentiate under the integral symbol"; i.e., that doing something like this (if we put $z = x + i y$), for example, $$ {\partial \over \partial x} f(z) = -{1 \over \pi} \int_{t = 0}^{t = \infty} \int_{\theta = 0}^{\theta = 2\pi} \; {\partial \over \partial x} \phi \left(z + e^{-t \theta}\right) \; d\theta \; dt $$ is legitimate.
Why do we need the compactness of support? To make sure that the integration over the unbounded domain $0 \leq t < \infty$ converges. Without this compactness, if we took $\phi$ to be constantly equal to $1$, the double integral would not exist.
Zorich's Математический анализ (Mathematical Analysis) may give a cleaner, more detailed and better paced explanation of differentiation under the integral.