[proofplan]
We compute the moments directly from the density of the standard semicircle law. Odd moments vanish by parity: the density is even and the monomial $x^{2n+1}$ is odd. For even moments, we reduce the integral over $[-2,2]$ to a beta integral by the substitutions $x = 2t$ and $u = t^2$, then evaluate the beta function using the beta-gamma identity and the half-integer values of the gamma function.
[/proofplan]
[step:Express the moments as integrals against the semicircle density]
For each integer $m \geq 0$, the definition of the distribution of $s$ gives
\begin{align*}
\varphi(s^m) = \int_{\mathbb{R}} x^m \rho(x)\,d\mathcal{L}^1(x).
\end{align*}
Since $\rho(x) = 0$ for $x \notin [-2,2]$, this becomes
\begin{align*}
\varphi(s^m) = \frac{1}{2\pi}\int_{-2}^{2} x^m \sqrt{4-x^2}\,d\mathcal{L}^1(x).
\end{align*}
The integrand is continuous on the compact interval $[-2,2]$, so the integral is finite for every $m \geq 0$.
[/step]
[step:Use parity to prove the odd moments vanish]
Fix an integer $n \geq 0$. Define
\begin{align*}
f_n: [-2,2] &\to \mathbb{R}
\end{align*}
\begin{align*}
x &\mapsto x^{2n+1}\sqrt{4-x^2}.
\end{align*}
The function $x \mapsto \sqrt{4-x^2}$ is even on $[-2,2]$, while $x \mapsto x^{2n+1}$ is odd. Hence $f_n$ is odd. Since $f_n$ is continuous on the symmetric interval $[-2,2]$, the change of variables $x = -y$ gives
\begin{align*}
\int_{-2}^{2} f_n(x)\,d\mathcal{L}^1(x) = 0.
\end{align*}
Therefore
\begin{align*}
\varphi(s^{2n+1}) = \frac{1}{2\pi}\int_{-2}^{2} x^{2n+1}\sqrt{4-x^2}\,d\mathcal{L}^1(x) = 0.
\end{align*}
[guided]
Fix an integer $n \geq 0$. The odd moment is
\begin{align*}
\varphi(s^{2n+1}) = \frac{1}{2\pi}\int_{-2}^{2} x^{2n+1}\sqrt{4-x^2}\,d\mathcal{L}^1(x).
\end{align*}
The factor $x^{2n+1}$ changes sign under $x \mapsto -x$, while the factor $\sqrt{4-x^2}$ does not. More formally, define
\begin{align*}
f_n: [-2,2] &\to \mathbb{R}
\end{align*}
\begin{align*}
x &\mapsto x^{2n+1}\sqrt{4-x^2}.
\end{align*}
Then, for every $x \in [-2,2]$,
\begin{align*}
f_n(-x) = (-x)^{2n+1}\sqrt{4-(-x)^2} = -x^{2n+1}\sqrt{4-x^2} = -f_n(x).
\end{align*}
Thus $f_n$ is odd. Because $f_n$ is continuous on $[-2,2]$, it is Lebesgue integrable. Splitting the integral at $0$ and applying the substitution $x = -y$ on $[-2,0]$, with $d\mathcal{L}^1(x)$ transformed to $d\mathcal{L}^1(y)$ and the interval $[-2,0]$ transformed to $[0,2]$, gives
\begin{align*}
\int_{-2}^{0} f_n(x)\,d\mathcal{L}^1(x) = \int_0^2 f_n(-y)\,d\mathcal{L}^1(y).
\end{align*}
Since $f_n(-y) = -f_n(y)$, we obtain
\begin{align*}
\int_{-2}^{0} f_n(x)\,d\mathcal{L}^1(x) = -\int_0^2 f_n(y)\,d\mathcal{L}^1(y).
\end{align*}
Adding the integral over $[0,2]$ gives
\begin{align*}
\int_{-2}^{2} f_n(x)\,d\mathcal{L}^1(x) = 0.
\end{align*}
Substituting the definition of $f_n$ into the moment formula therefore proves
\begin{align*}
\varphi(s^{2n+1}) = 0.
\end{align*}
[/guided]
[/step]
[step:Reduce the even moment integral to a beta integral]
Fix an integer $n \geq 0$. From the density formula,
\begin{align*}
\varphi(s^{2n}) = \frac{1}{2\pi}\int_{-2}^{2} x^{2n}\sqrt{4-x^2}\,d\mathcal{L}^1(x).
\end{align*}
Apply the substitution $x = 2t$. The interval $[-2,2]$ becomes $[-1,1]$, and one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) transforms by $d\mathcal{L}^1(x) = 2\,d\mathcal{L}^1(t)$. Therefore
\begin{align*}
\varphi(s^{2n}) = \frac{2^{2n+1}}{\pi}\int_{-1}^{1} t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t).
\end{align*}
The integrand $t \mapsto t^{2n}(1-t^2)^{1/2}$ is even, so
\begin{align*}
\int_{-1}^{1} t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t) = 2\int_0^1 t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t).
\end{align*}
Now apply the substitution $u = t^2$ on $[0,1]$. The interval $[0,1]$ maps to $[0,1]$, and $d\mathcal{L}^1(t) = \frac{1}{2}u^{-1/2}\,d\mathcal{L}^1(u)$. Hence
\begin{align*}
2\int_0^1 t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t) = \int_0^1 u^{n-\frac{1}{2}}(1-u)^{\frac{1}{2}}\,d\mathcal{L}^1(u).
\end{align*}
Define the beta function $B: (0,\infty)\times(0,\infty) \to (0,\infty)$ by
\begin{align*}
B(a,b) = \int_0^1 u^{a-1}(1-u)^{b-1}\,d\mathcal{L}^1(u).
\end{align*}
Thus
\begin{align*}
\varphi(s^{2n}) = \frac{2^{2n+1}}{\pi}B\left(n+\frac{1}{2},\frac{3}{2}\right).
\end{align*}
[guided]
We now compute the even moment. The starting point is the same density formula:
\begin{align*}
\varphi(s^{2n}) = \frac{1}{2\pi}\int_{-2}^{2} x^{2n}\sqrt{4-x^2}\,d\mathcal{L}^1(x).
\end{align*}
The purpose of the substitution $x = 2t$ is to normalize the support interval from $[-2,2]$ to $[-1,1]$. Under this substitution, $x^{2n} = 2^{2n}t^{2n}$, $\sqrt{4-x^2} = 2(1-t^2)^{1/2}$, and one-dimensional Lebesgue measure transforms as $d\mathcal{L}^1(x) = 2\,d\mathcal{L}^1(t)$. The domain $[-2,2]$ becomes $[-1,1]$. Therefore
\begin{align*}
\varphi(s^{2n}) = \frac{1}{2\pi}\int_{-1}^{1} 2^{2n}t^{2n}\,2(1-t^2)^{1/2}\,2\,d\mathcal{L}^1(t).
\end{align*}
Collecting the constant factors gives
\begin{align*}
\varphi(s^{2n}) = \frac{2^{2n+1}}{\pi}\int_{-1}^{1} t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t).
\end{align*}
The function $t \mapsto t^{2n}(1-t^2)^{1/2}$ is even on $[-1,1]$, so the integral over the symmetric interval is twice the integral over $[0,1]$:
\begin{align*}
\int_{-1}^{1} t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t) = 2\int_0^1 t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t).
\end{align*}
The remaining integral has the shape of a beta integral after the substitution $u = t^2$. On $[0,1]$, this substitution maps $[0,1]$ onto $[0,1]$, and the one-dimensional change-of-variables formula gives
\begin{align*}
d\mathcal{L}^1(t) = \frac{1}{2}u^{-1/2}\,d\mathcal{L}^1(u).
\end{align*}
Also $t^{2n} = u^n$ and $(1-t^2)^{1/2} = (1-u)^{1/2}$. Hence
\begin{align*}
2\int_0^1 t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t) = 2\int_0^1 u^n(1-u)^{1/2}\frac{1}{2}u^{-1/2}\,d\mathcal{L}^1(u).
\end{align*}
After simplifying the powers of $u$, this becomes
\begin{align*}
2\int_0^1 t^{2n}(1-t^2)^{1/2}\,d\mathcal{L}^1(t) = \int_0^1 u^{n-\frac{1}{2}}(1-u)^{\frac{1}{2}}\,d\mathcal{L}^1(u).
\end{align*}
Define the beta function $B: (0,\infty)\times(0,\infty) \to (0,\infty)$ by
\begin{align*}
B(a,b) = \int_0^1 u^{a-1}(1-u)^{b-1}\,d\mathcal{L}^1(u).
\end{align*}
With $a = n+\frac{1}{2}$ and $b = \frac{3}{2}$, the last integral is exactly $B\left(n+\frac{1}{2},\frac{3}{2}\right)$. Thus
\begin{align*}
\varphi(s^{2n}) = \frac{2^{2n+1}}{\pi}B\left(n+\frac{1}{2},\frac{3}{2}\right).
\end{align*}
[/guided]
[/step]
[step:Evaluate the beta integral using gamma function identities]
Define the gamma function $\Gamma: (0,\infty) \to (0,\infty)$ by
\begin{align*}
\Gamma(r) = \int_0^\infty y^{r-1}e^{-y}\,d\mathcal{L}^1(y).
\end{align*}
We use the classical beta-gamma identity
\begin{align*}
B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
\end{align*}
for $a,b > 0$; this is a standard special-function identity not yet cited from the wiki. Here $a = n+\frac{1}{2} > 0$ and $b = \frac{3}{2} > 0$, so
\begin{align*}
B\left(n+\frac{1}{2},\frac{3}{2}\right) = \frac{\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma(n+2)}.
\end{align*}
The half-integer gamma values are
\begin{align*}
\Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}
\end{align*}
and
\begin{align*}
\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{4^n n!}.
\end{align*}
Also $\Gamma(n+2) = (n+1)!$. Substituting these values gives
\begin{align*}
\varphi(s^{2n}) = \frac{2^{2n+1}}{\pi}\frac{\frac{(2n)!\sqrt{\pi}}{4^n n!}\frac{\sqrt{\pi}}{2}}{(n+1)!}.
\end{align*}
Since $4^n = 2^{2n}$, the constants simplify to
\begin{align*}
\varphi(s^{2n}) = \frac{(2n)!}{n!(n+1)!}.
\end{align*}
Finally,
\begin{align*}
\frac{(2n)!}{n!(n+1)!} = \frac{1}{n+1}\frac{(2n)!}{n!n!} = \frac{1}{n+1}\binom{2n}{n} = C_n.
\end{align*}
Therefore $\varphi(s^{2n}) = C_n$ for every integer $n \geq 0$.
[/step]
[step:Combine the parity and beta computations]
For every integer $n \geq 0$, the parity argument gives
\begin{align*}
\varphi(s^{2n+1}) = 0,
\end{align*}
and the beta-gamma evaluation gives
\begin{align*}
\varphi(s^{2n}) = \frac{1}{n+1}\binom{2n}{n} = C_n.
\end{align*}
This proves the stated Catalan moment formula for the standard semicircle law.
[/step]
