Calculate: $\int_{\mathbb{D}}\sum_{k=0}^s{s \choose k}\bar{z}^sz^{2k}d\mu(z)$ $Problem:$ Let $\mathbb{D}:={\{z\in\mathbb{C}:|z|<1}\}$ the unit open disk in the complex plane, and let $ \mu $ be the Lebesgue measure in $\mathbb{D}$, calculate:
$$\int_{\mathbb{D}}\sum_{k=0}^s{s \choose k}\bar{z}^sz^{2k}d\mu(z)$$
$Idea:$ Note that being a finite series, we can exchange the integral with the series, so: $$\int_{\mathbb{D}}\sum_{k=0}^s{s \choose k}\bar{z}^sz^{2k}d\mu(z)=\sum_{k=0}^s{s \choose k}\int_{\mathbb{D}}\bar{z}^sz^{2k}d\mu(z)$$ How can you continue? Polar coordinates? Welcome any suggestion. Thank you.
 A: Rewrite the integral in the following way
$$\int_{\mathbb{D}}\sum_{k=0}^s{s \choose k}\bar{z}^sz^{2k}d\mu(z) = \int_{\mathbb{D}}\bar{z}^s(1+z^2)^sd\mu(z)$$
Then use the following theorem from complex analysis
$$2i\int_{D}\frac{\partial f}{\partial \bar{z}}\:d\mu(z) = \int_{\partial D}f(z)\:dz$$
to turn the integral into a line integral like so:
$$\int_{\mathbb{D}}\bar{z}^s(1+z^2)^sd\mu(z) = \frac{1}{2i(s+1)}\int_{\partial \mathbb{D}} \bar{z}^{s+1}(1+z^2)^s \:dz = \frac{1}{2i(s+1)}\int_{\partial \mathbb{D}} \frac{(1+z^2)^s}{z^{s+1}} \:dz$$
since on $\partial \mathbb{D}$ we have that $\bar{z} = z^{-1}$. Then from Cauchy's integral formula we have that
$$\frac{1}{2i(s+1)}\int_{\partial \mathbb{D}} \frac{(1+z^2)^s}{z^{s+1}} \:dz = \frac{1}{2i(s+1)}\left(\frac{2\pi i}{s!} \frac{d^s}{dz^s}(1+z^2)^s\right)\Biggr|_{z=0}$$
$$= \frac{\pi}{(s+1)!}\frac{d^s}{dz^s}(1+z^2)^s\Biggr|_{z=0} = \frac{\pi}{(s+1)!}\frac{d^s}{dz^s}\sum_{k=0}^s{s \choose k}z^{2k}\Biggr|_{z=0}$$
from Taylor's theorem we also have that
$$f^{(n)}(a) = c_n\cdot n!$$
where $c_n$ is the coefficient of the $n$th power of the function's Taylor series. Notice that in the series, that occurs at $k = \frac{s}{2}$, giving us
$$\frac{\pi}{(s+1)!}\frac{d^s}{dz^s}\sum_{k=0}^s{s \choose k}z^{2k}\Biggr|_{z=0} = \frac{\pi}{(s+1)!}\left({s \choose \frac{s}{2}}\cdot s!\right) = \frac{\pi}{s+1} {s \choose \frac{s}{2}}$$
when $s$ is even and $0$ when $s$ is odd.
