$Problem:$ 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:
$$\int_{\mathbb{D}}\left(\sum_{k=0}^s{s \choose k}|z|^{2k}\right)^2d\mu(z)$$
$ Idea: \textbf{According to the comments}$ By the binomial theorem $ \sum_ {k = 0} ^ s {s \choose k} | z | ^ {2k} = (1+ | z | ^ 2) ^ s $. Then: $$ \int _ {\mathbb {D}} \left (\sum_ {k = 0} ^ s {s \choose k} | z | ^ {2k} \right) ^ 2d \mu (z) = \int _ {\mathbb {D}} (1+ | z | ^ 2) ^ {2s} $$ Thus using polar coordinates, $ z = re ^ {i \theta} $, with $ 0 \leq r = | z | \leq 1 $ and $ \theta \in [0,2 \pi] $ then: $$ \int _ {\mathbb {D}} (1+ | z | ^ 2) ^ {2s} = \int_ {0 } ^ {1} \int_ {0} ^ {2 \pi} (1 + r ^ 2) ^ {2s} rdrd \theta $$ $$= \pi \int_ {0} ^ {1} (1 + r ^ 2) ^ {2s} 2rdr = \pi \int_ {1} ^ {2} w ^ {2s} dw = \pi (\frac {2 ^ {2s + 1} -1} {2s + 1}) $$
Is my solution correct?. Thank you for reading.