Dumbbell Contour? $\int_0^1 \log(x)\log(1-x)dx$ via complex methods. Having evaluated this integral via the power series and various approaches via special functions, I'm now curious if there is a direct way to compute this integral by taking a slit along $[0,1]$ and using a 'dogbone' contour with the residue at infinity to obtain the famous result. My own attempts haven't led to much progress.
Please note: I have seen the many threads related to this integral, but not one that has approached the integral in this way so I hope this does not get flagged up as a duplicate.
 A: The reason that the dog bone contour is inapplicable here is that branch cuts from, say, $0$ to $\infty$ and $1$ to $\infty$ do not "collapse" into the "slit" from $0$ to $1$.
To see this, we note that for $z=x+iy$, $x>1$ and $y\to 0+$ we have $\arg(z)=0$ and $\arg(1-z)=-\pi$, while for $x>1$ and $y\to 0^-$, we have $\arg(z)=2\pi$ and $\arg(z)=\pi$.
Therefore, on the upper part of the coalescing branch cuts for which $x>1$
$$\begin{align}
f(z)&=\log(z)\log(1-z)\\\\
&=(\log(x)+i0)(\log(|1-x|)-i\pi)\\\\
&=\log(x)\log(|1-x|)-i\pi \log(x)
\end{align}$$ 
while on the lower part of the coalescing branch cuts for which $x>1$
$$\begin{align}
f(z)&=\log(z)\log(1-z)\\\\
&=(\log(x)+i2\pi)(\log(|1-x|)+i\pi)\\\\
&=\log(x)\log(|1-x|)+i\pi \log(x)+i2\pi \log(|1-x|)-2\pi^2
\end{align}$$ 

Therefore, the function $f(z)$ is not continuous across the coalescing branch cuts and they do not, therefore, collapse into a "slit."


NOTE:
This situation is different from the case for which $f(z)=z^{1/2}(1-z)^{1/2}$.  Following the preceding analysis, we find that on the upper part of the  coalescing branch cuts 
$$\begin{align}
f(z)&=\sqrt{z} \sqrt{1-z} \\\\
&=\sqrt{x}\sqrt{x-1}e^{-i\pi/2}\\\\
&=-i\sqrt{x}\sqrt{x-1}
\end{align}$$ 
while on the lower part of the coalescing branch cuts for which $x>1$
$$\begin{align}
f(z)&=\sqrt{z}\sqrt{1-z}\\\\
&=\sqrt{x}e^{i\pi}\sqrt{x-1}e^{i\pi/2}\\\\
&=-i\sqrt{x}\sqrt{x-1}
\end{align}$$ 

Therefore, the function $f(z)$ is continuous across the coalescing branch cuts and they do, therefore, collapse into a "slit."

