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:


$\int\_{0}^{n} [C\_n(x)\cdot f(y)^x \cdot g(z)^{n-x}dx] = [f(y)+g
(z)]^{n}$




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Another integral: let $f,g : R \mapsto R$


Is $\int\_{-\infty}^{\infty}\int\_{-y}^{y} f(y-x)g(y+x)dxdy = [\int\_{-\infty}^{\infty}f(x)dx][\int\_{-\infty}^{\infty}g(x)dx]$



Integrals I am curious about
We have the classical [Euler integral](https://en.wikipedia.org/wiki/Beta_function)


$$B(x+1,n-x+1):=\int\_0^1y^x(1-y)^{n-x} dy=\frac{\Gamma(x+1)\Gamma(n-x+1)}{\Gamma(n+2)} $$
$$
=\frac{x(n-x)}{n+1}\frac{\Gamma(x)\Gamma(n-x)}{\Gamma(n+1)}=\frac{x(n-x)}{(n+1)C\_n(x)}
$$
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 $1$


