is it possible to prove the identity $$\forall x > 0,\ \forall y > 0,\quad B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ where $\Gamma$ is the Gamma function and $B$ is the Beta function without using the Fubini theorem (the standard proof using it and polar coordinates)?
It am looking for an elementary proof of this identity...
We can prove the identity in a different way using Euler's infinite product representation of the gamma function,
$$\Gamma(x) = \lim_{n \to \infty}\Gamma_n(x) = \lim_{n \to \infty} \frac{n!n^x}{(x+1)(x+2) \cdots (x+n)}$$
The beta function defined as $\displaystyle B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt$ satisfies the recurrence relation
$$ \tag{1}B(x,y) = \frac{x+y}{y}B(x,y+1)$$
The proof of (1) involves integration by parts and can be furnished if desired. Applying (1) $n$ times we get
$$\tag{2}B(x,y) = \frac{(x+y)(x+y+1) \cdots (x + y + n)}{y(y+1) \cdots (y +n)}B(x,y+n+1) \\ =\frac{\Gamma_n(y)}{\Gamma_n(x+y)}n^{x}B(x,y+n+1)$$
Note that
$$n^{x}B(x,y+n+1) = n^{x}\int_0^1t^{x-1}(1-t)^{y+n} \, dt,$$
and after changing variables with $s = nt$ we get
$$\tag{3}n^{x}B(x,y+n+1) = \int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{y+n} \, ds = \int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds $$
Substituting into (2) using (3) yields
$$\tag{4}B(x,y) = \frac{\Gamma_n(y)}{\Gamma_n(x+y)}\int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds$$
By the dominated convergence theorem,
$$\lim_{n \to \infty}\int_0^ns^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}\, ds = \lim_{n \to \infty}\int_0^{\infty}s^{x-1}\left(1-\frac{s}{n}\right)^{n} \left(1-\frac{s}{n}\right)^{y}1_{[0,n]}\, ds \\ = \int_0^\infty s^{x-1}e^{-s}\, ds ,$$
and, since $\lim_{n \to \infty} \Gamma_n(z) = \Gamma(z)$, upon taking the limit of both sides of (4) as $n \to \infty$ we obtain
$$B(x,y) = \frac{\Gamma(y)}{\Gamma(x+y)}\int_0^\infty s^{x-1}e^{-s}\, ds = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$