How can I justify that I can put the laplace operator under the integral?

Consider the integral:

h(r,θ)=12π∫2π0g(ϕ)1−r21−2rcos(θ−ϕ)+r2dϕ,r<1$h(r,\theta )=\frac{1}{2\pi }{\int }_0^{2\pi }g(\varphi )\frac{1-r^2}{1-2r\cos (\theta -\varphi )+r^2}d\varphi ,r<1$

I want to show that Δh=0$\mathrm{\Delta }h=0$, but in order to do so, I need to justify this:

Δ12π∫2π0g(ϕ)1−r21−2rcos(θ−ϕ)+r2dϕ=12π∫2π0g(ϕ)Δ1−r21−2rcos(θ−ϕ)+r2dϕ$\mathrm{\Delta }\frac{1}{2\pi }{\int }_0^{2\pi }g(\varphi )\frac{1-r^2}{1-2r\cos (\theta -\varphi )+r^2}d\varphi =\frac{1}{2\pi }{\int }_0^{2\pi }g(\varphi )\mathrm{\Delta }\frac{1-r^2}{1-2r\cos (\theta -\varphi )+r^2}d\varphi$

That way, this becomes 0$0$, as the Poisson kernel is harmonic.