How do I see if $g$ is a polynomial or not?? Let $u$ be a real valued harmonic function on $\mathbb{C}$. Let $g: \mathbb{R^2} \to \mathbb{R}$ be defined by:
$$g(x,y)=\int_0^{2\pi}u(e^{i\theta}(x+iy))\sin \theta \,d\theta$$
Which of the following statements is TRUE?
(A) $$ is a harmonic polynomial
(B) $$ is a polynomial but not harmonic
(C) $$ is harmonic but not a polynomial
(D) $$ is neither harmonic nor a polynomial
Differentiating under the integral sign I see that $g$ is harmonic. But How do I conclude if it is a polynomial or not??
 A: Since $u$ is real valued and harmonic on $\mathbb C,$ $u = \text { Re } f$ for some entire function $f.$ We can write $f(z)=\sum_{n=0}^{\infty}a_n z^n,$ the power series converging uniformly on compact subsets of $\mathbb C.$ Then
$$g(z) = \text { Re }\int_0^{2\pi}f(e^{i\theta}z)\sin \theta\, d \theta
= \text { Re } \int_0^{2\pi}\left (\sum_{n=0}^{\infty}a_n(e^{i\theta}z)^n\right )\frac{e^{i\theta} - e^{-i\theta} }{2i}\, d \theta
$$ $$=\text { Re }\frac{1}{2i}\sum_{n=0}^{\infty}a_nz^n\int_0^{2\pi}e^{in\theta}(e^{i\theta} - e^{-i\theta} )\, d \theta.$$
The interchange of integral and summation is justified by uniform convergence. The integrals in the last expression are $0$ except for $n=1,$ where the integral equals $-2\pi.$ Thus
$$g(z) = \text { Re}\left (-\frac{\pi}{i}a_1 z\right).$$
Therefore $g$ is the real part of a holomorphic polynomial of degree $1,$ and it follows that $g$ is a real harmonic polynomial of degree $1.$
A: $$g(x,y) = \int\limits_0^{2\pi}u(e^{i\theta}(x+iy))\sin\theta d\theta = \int\limits_0^{2\pi}u(e^{i\theta}(x+iy))\dfrac{e^{i\theta} - e^{-i\theta}}{2i}\dfrac {de^{i\theta}}{ie^{i\theta}}$$
$$ = -\dfrac12\oint\limits_{|z|=1}u(z(x+iy))\left(1-\dfrac1{z^2}\right)\,dz.$$
In accordance with conditions, $u(x)$ is real valued, continuous and differentiable function.
Using residue theorem,
$$g(x,y) = \dfrac12\cdot2\pi i\cdot\mathop{Res}\limits_{z=0}\dfrac{u(z(x+iy))}{z^2} = \pi i\lim\limits_{z\to 0}\dfrac d{dz}u(z(x+iy)) = $$
$$\pi i(x+iy)\lim\limits_{z\to 0}u'(z(x+iy)) = \pi i(x+iy)u'(0).$$
So $g(x, y)$ is polynomial.
