Show that $\int_{0}^{1}\int_{0}^{1}{x\over (k-xy)\ln(xy)}\mathrm\, dx\,dy=(k-1)\ln\sigma_k?$ Experimental shown that

$$\int_{0}^{1}\int_{0}^{1}{x\over (k-xy)\ln(xy)}\mathrm \,dx\,dy=(k-1)\ln(\sigma_k)\tag1$$
  $k\ge1$

where $$\sigma_k=1^{1\over k}\cdot 2^{1\over k^2} \cdot 3^{1\over k^3} \cdot 4^{1\over k^4} \cdots$$
How can we prove $(1)$?
 A: 
There is a sign error in the OP.  The result must be non-positive for $k>1$.  In fact, we have for $k>1$
$$\bbox[5px,border:2px solid #C0A000]{\int_0^1\int_0^1 \frac{x}{(k-xy)\log(xy)}\,dx\,dy=(1-k)\log(\sigma_k)}\tag 1$$
as we shall proceed to show. 


Let $I(k)$ be given by the integral 
$$I(k)=\int_0^1\int_0^1 \frac{x}{(k-xy)\log(xy)}\,dx\,dy\tag2$$
for $k>1$.

Interchanging the order of integration and letting $y=t/x$ in $(2)$, we find that 
$$I(k)=\int_0^1 \int_0^x \frac{1}{(k-t)\log(t)}\,dt\,dx \tag3$$

Next, integrating by parts the outer integral in $(3)$ with $u=\int_0^x \frac{1}{(k-t)\log(t)}\,dt$ and $v=x$ reveals
$$\begin{align}
I(k)&=\int_0^1 \frac{1}{(k-t)\log(t)}\,dt-\int_0^1 \frac{x}{(k-x)\log(x)}\,dx\\\\
&=-\int_0^1 \frac{x-1}{(k-x)\log(x)}\,dx\\\\
&=-\int_0^1 \frac{1}{k-x}\int_0^1 x^s\,ds\,dx\\\\
&=-\frac1k \int_0^1 \frac{1}{1-\frac xk}\int_0^1 x^s\,ds\,dx\\\\
&=-\sum_{n=0}^\infty \frac1{k^{n+1}} \int_0^1\int_0^1 x^{s+n}\,dx\,ds\\\\
&=-\sum_{n=0}^\infty \frac{1}{k^{n+1}}\log\left(\frac{n+2}{n+1}\right)\\\\
&=-\sum_{n=0}^\infty \frac{1}{k^{n+1}}\left( \log(n+2)-\log(n+1)\right)\\\\
&=(1-k)\sum_{n=1}^\infty \frac{\log(n+1)}{k^{n+1}}\\\\
&=(1-k)\sum_{n=2}^\infty \frac{\log(n)}{k^{n}}\\\\
&=(1-k)\log(\sigma_k)
\end{align}$$

Therefore, we find

$$\bbox[5px,border:2px solid #C0A000]{\int_0^1\int_0^1 \frac{x}{(k-xy)\log(xy)}\,dx\,dy=(1-k)\log(\sigma_k)}$$

which agrees with $(1)$!
