Let $C_n(x) = \frac{n!}{\Gamma(x)\cdot \Gamma(n-x)}$
Is the following true: $\int_{0}^{n} [C_n(x)\cdot y^x \cdot (1-y)^{n-x}dx] = 1$??
just wondering
In generality for continuous functions $f,g$ from the reals to the reals is it the case that:
$\int_{0}^{n} [C_n(x)\cdot f(y)^x \cdot g(z)^{n-x}dx] = [f(y)+g (z)]^{n}$
Another integral: let $f,g : R \mapsto R$
Is $\int_{-\infty}^{\infty}\int_{-y}^{y} f(y-x)g(y+x)dxdy = [\int_{-\infty}^{\infty}f(x)dx][\int_{-\infty}^{\infty}g(x)dx]$
We have the classical Euler integral
$$B(x+1,n-x+1):=\int_0^1y^x(1-y)^{n-x} dy=\frac{\Gamma(x+1)\Gamma(n-x+1)}{\Gamma(n+2)} $$ $$ =\frac{x(n-x)}{n+1}\frac{\Gamma(x)\Gamma(n-x)}{\Gamma(n+1)}=\frac{x(n-x)}{(n+1)C_n(x)} $$ Hence $$ \int_0^1\underbrace{C_n(x)y^x(1-y)^{n-x}}_{=:f_n(x,y)}\;dy =\frac{x(n-x)}{n+1}. $$ You are asking if $$ I_n(y):=\int_0^n f_n(x,y) dx=1,\;\;\forall y\in[0,1]. $$ We have $$ \int_0^1I(y) dy=\int_0^1 dy\int_0^nf_n(x,y) dx=\int_0^ndx\int_0^1f_n(x,y) dy $$ $$ =\frac{1}{n+1}\int_0^nx(n-x) dx=\frac{n^3}{6(n+1)}. $$ Thus $$ \int_0^1I_n(y)\, dy=\frac{n^3}{6(n+1)}. $$ In particular, $I_n(y)$ cannot be identically equal to $1$
