Determine the convergence and divergence of $\int_0^1{\int_0^1{\frac{1}{x^{1-y}\left( 1-x \right) ^y}dxdy}}$ Determine whether $$\int_0^1{\int_0^1{\frac{1}{x^{1-y}\left( 1-x \right) ^y}\mathbb dx \mathbb dy}}$$
is convergent or divergent.
It looks like that it needs to be scaled by inequalities, but I failed. Can anyone help?
 A: Not the simplest argument perhaps, but the inner integral (over $x$) is well-known: $$\int_0^1\frac{dx}{x^{1-y}(1-x)^y}=\mathrm{B}(y,1-y)=\Gamma(y)\Gamma(1-y)=\frac{\pi}{\sin\pi y},$$ and since $\int_0^1\frac{dy}{\sin\pi y}$ diverges, the original integral does as well (the integrand is positive).
A: $$I=\int_0^1\int_0^1\frac{1}{x^{1-y}(1-x)^y}dxdy=\int_0^1\int_0^1x^{y-1}(1-x)^{-y}dxdy$$
now using the definition of the Beta function we can say:
$$I=\int_0^1B(y,1-y)dy$$
Now we also have that:
$$B(y,1-y)=\frac{\pi}{\sin(\pi y)}\,\,\,\,y\notin\mathbb{Z}$$
so your integral becomes:
$$I=\pi\int_0^1\csc(\pi y)dy$$
$z=\pi y\Rightarrow dy=\frac{dz}\pi$
$$I=\int_0^\pi\csc(z)dz$$
Which you can see is divergent due to the values of $\csc(z)$ at $z=0,\pi$
A: Here is an answer that avoid the use of beta function:
$$ I
= \int_{0}^{1} \frac{1}{x} \int_{0}^{1} \left(\frac{x}{1-x}\right)^y \, \mathrm{d}y \mathrm{d}x
= \int_{0}^{1} \frac{2x-1}{x(1-x)\log\bigl(\frac{x}{1-x}\bigr)} \, \mathrm{d}x. $$
Near $x = 0$, the integrand is asymptotically
$$ \frac{2x-1}{x(1-x)\log\bigl(\frac{x}{1-x}\bigr)} \sim -\frac{1}{x\log x}, $$
hence the integral diverges by the limit comparison test with
$$ - \int_{0}^{\frac{1}{2}} \frac{1}{x\log x} \, \mathrm{d}x = \biggl[ -\log(-\log x) \biggr]_{0}^{\frac{1}{2}} = \infty. $$
