How to prove $\int_0^{2\pi} f(a+rcos(x),b+rsin(x)) dx =2{\pi}f(a,b)$?

Question

Recently I've been looking around for integral tricks here, here, and here (just to name a few). I came across the post on brilliant's website here (move the slider/ scroll down to the very bottom to see the information) with a really nice formula for integrals of the form:

$\int_0^{2\pi} f(a+rcos(x),b+rsin(x)) dx =2{\pi}f(a,b)$

Initially I thought this could be related to cauchy's integral formula:

$f(a)=\frac{1}{2{\pi}i}\int_{\gamma}\frac{f(z)}{z-a}dz$

Which can be transformed via the substitution $z=a+re^{i\theta}:\theta {\in}[0,2{\pi}]$ to:

$f(a)=\frac{1}{2{\pi}}\int_0^{2\pi}f(a+re^{i\theta})d{\theta}$

Thus:

$2{\pi}f(a)=\int_0^{2\pi}f(a+re^{i\theta})d{\theta}$

This is the result from where I ask for help and understanding as it looks as though there might be a way to prove this from here but I don't know what I'm doing necessarily when it comes to multivariable functions such as the one I would like to solve for and if this actually could lead to the desired result. Thank you, much appreciated.

Answer 1

Thank you Martin R your insight was very useful!. I did some further research on the mean value theorem for harmonic functions that you mentioned and I found a video here which actually goes over the very equation and derivation of the mean value theorem that's used!.
as seen in the video:

$0=\int_{\gamma}u_xdy-u_ydx$

[$r(t)=(x(t),y(t))=(a+rcos(t),b+rsin(t)): 0{\le}t{\le}{2\pi}$

[$r'(t)=(x'(t),y'(t))=(-rsin(t),rcos(t))$

$0=\int_0^{2 \pi}u_x(a+rcos(t),b+rsin(t))rcos(t)-u_y(a+rcos(t),b+rsin(t))(-rsin(t))dt$

$0=\int_0^{2 \pi}\frac{\partial}{\partial r}(u(a+rcos(t),b+rsin(t)))dt$

$0=\frac{d}{dr}\int_0^{2 \pi}u(a+rcos(t),b+rsin(t))dt$

$0=\int_0^r\frac{d}{dr}\int_0^{2 \pi}u(a+rcos(t),b+rsin(t))dtdr$

$0=\int_0^{2 \pi}u(a+rcos(t),b+rsin(t))dt-\int_0^{2 \pi}u(a,b)dt$

so clearly,
${2\pi}u(a,b)=\int_0^{2 \pi}u(a+rcos(t),b+rsin(t))dt$ given that 'u' (in this case) is a harmonic function.