[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*}[/step]

[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: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*}[/step]

[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: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]