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

Consider the integral:

$$h(r, \theta) = \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi, r<1$$

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

$$\Delta \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi = \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \Delta \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi$$

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

• There's something not right about the integral, I think you have your letters mixed up – Ninad Munshi Jul 24 '20 at 3:30
• probably $d\phi$ so that its a convolution integral – Calvin Khor Jul 24 '20 at 3:36
• Whoops,... yeah it's $d\phi$ – Kay Jul 24 '20 at 3:36
• It looks a bit like the Leibniz integral rule. It's just the $\phi$-component of $\Delta$ which seems a bit fishy to pull into the integral. – Vercassivelaunos Jul 24 '20 at 10:14