Show u=∈t_0^(2π)f(xcos t+ysin t+iz, t) dt solves the 3D…
Show $u=\int_0^{2\pi}f(x\cos t+y\sin t+iz,\,t)\,dt$ solves the 3D Laplace equation.
Explanation of Method
Differentiate under the integral sign; the chain-rule coefficients $(\cos t,\sin t,i)$ satisfy the complex isotropic-cone identity $\cos^2t+\sin^2t+i^2=0$, which forces the Laplacian of the integrand to vanish identically.
Step 1 · Differentiate under the integral
With $\xi=x\cos t+y\sin t+iz$ (so all second partials of $\xi$ in $x,y,z$ vanish): \[ u_{xx}=\int_0^{2\pi}f_{\xi\xi}\cos^2t\,dt,\quad u_{yy}=\int_0^{2\pi}f_{\xi\xi}\sin^2t\,dt,\quad u_{zz}=\int_0^{2\pi}f_{\xi\xi}\,i^2\,dt. \]
Step 2 · Sum and use the isotropic identity
\[ \Delta u = \int_0^{2\pi}f_{\xi\xi}(\cos^2t+\sin^2t+i^2)\,dt = \int_0^{2\pi}f_{\xi\xi}(1-1)\,dt = 0. \]
Remark
The vector $(\cos t,\sin t,i)$ lies on the complex isotropic (null) cone $\lambda_1^2+\lambda_2^2+\lambda_3^2=0$ of the Laplace operator, which has no real solutions but a circle's worth of complex ones parametrised by $t$. Integrating an analytic function of $\xi$ over this family automatically yields a harmonic function — the classical Whittaker–Bateman representation of harmonic functions in three variables.
Result
Conclusion: