Integration of two dilogarithms: $\int_{0}^{\frac{1}{B_{1}}}\frac{dx}{x}\ln(x-1)+\int_{0}^{\frac{1}{B_{2}}}\frac{dx}{x}\ln(x-1)$ I have an integral:
$\int_{0}^{\frac{1}{B_{1}}}\frac{dx}{x}\ln(x-1)+\int_{0}^{\frac{1}{B_{2}}}\frac{dx}{x}\ln(x-1)$
where
$1<\frac{1}{B_{1}}\leq2 , 2\leq\frac{1}{B_{2}}$
I need to find some finite result. The result should look like 
~$(\ln\frac{B_{1}}{B_{2}}+i\pi)^{2}$ or something like that. I am not really sure. Please help. I am stuck with this for a really long time (weeks). Need some hints of complex analysis I think...
B1 and B2 are dependent on each other: $B_{12}=\frac{1\pm\sqrt{1-4\tau}}{2}$ , where $0<\tau<1/4$ . There must be the way to get a beautiful and finte answer. 
 A: If $0<x<1$, then, assuming the principal branch of the logarithm,
$$
\ln(x-1)=\ln|x-1|+\pi\,i.
$$
Since none of the integrals
$$
\int_0^1\frac{\ln|x-1|}{x}\,dx\ ,\quad\int_0^1\frac{1}{x}\,dx
$$
is convergent, the integrals in your question are not defined.
A: From a formal point of view we have :
$$f(z)=\int \frac{\log(z-1)}z dz=\log(z)\log(z-1)+\int \frac{\log(z)}{1-z}dz$$
$$f(z)=\log(z)\log(z-1)-\log(z)\log(1-z)-\int \frac{\log(1-z)}z dz$$
$$f(z)=\log(z)\left(\log(z-1)-\log(1-z)\right)-\mathrm{Li}_2(z)$$
( $\log(z-1)-\log(1-z)$ is simply $+\pi i$ for $\Im(z)>0$ and $-\pi i$ for $\Im(z)<0$ and we could have obtained the same result by integrating $\frac{\log(z-1)}z=\frac{\log\left((1-z)e^{\pm \pi i}\right)}z$ ).
Here are the pictures of the real and imaginary parts of this function :

Both parts are very smooth near $2$ but the behavior of the imaginary part near $0$ is divergent (because of the $\pm \log(z) \pi i$ part).
At this point I see only two ways to make sense of your formula $\displaystyle I(B_1,B_2)=\int_{0}^{\frac{1}{B_{1}}}\frac{dx}{x}\ln(x-1)+\int_{0}^{\frac{1}{B_{2}}}\frac{dx}{x}\ln(x-1)\ $ :


*

*subtract the integrals instead of adding them with the advantage of getting different signs for $B_1$ and $B_2$ (and possibly a term $\log\left(\frac{B_1}{B_2}\right)$). You should get :
$$I(B_1,B_2)=-\mathrm{Li}_2\left(\frac 1{B_1}\right)+\mathrm{Li}_2\left(\frac 1{B_2}\right)$$

*keep your addition of integrals but 'shift' their paths in the complex plane for example by $+i\epsilon$ and $-i\epsilon$ so that both 'regularized' infinite contributions will cancel at $0$ and your sum will become simply (neglecting the $\epsilon$ term for $z \approx 2$ by replacing $\mathrm{Li}_2(z\pm i \epsilon)$ with $\mathrm{Li}_2(z)$) :
$$I(B_1,B_2)=-\mathrm{Li}_2\left(\frac 1{B_1}\right)-\mathrm{Li}_2\left(\frac 1{B_2}\right)$$
In this case :
$$I\left(\frac 12,\frac 12\right)=-2\mathrm{Li}_2(2)=-\frac{\pi^2}2+2\pi i\log(2)= \frac{\pi^2}2-\log(2)^2+\left(\log(2)+i\pi\right)^2$$ 
looks a little like you wished but closed forms for dilogarithms are sparse (see functions.Wolfram).
In your special case $B_{12}=\frac{1\pm\sqrt{1-4\tau}}{2}$ (so that $B_1\cdot B_2=\tau$) I got these special values :  
$
\begin{array} {c|l}
 \tau & I(B_1,B_2) \\
\hline \\
0 & -\frac{\pi^2}6 \\
\frac 14 & -\frac{\pi^2}6+\log(2)^2 \\ 
\frac 12 & -5\frac{\pi^2}{48}+\frac{\log(2)^2}4 \\ 
1 & -\frac{\pi^2}{18}\\
\end{array}
$
UPDATE: Since $B_1+B_2=1$ let's use one of the dilogarithm identities to get a closed form (for $z>1$) :
$$-\mathrm{Li}_2\left(\frac 1{\frac 1z}\right)-\mathrm{Li}_2\left(\frac 1{1-\frac 1z}\right)=\frac{\log(z-1)^2}2-\frac{\pi^2}2-i\pi\log\left(\frac{z-1}{z^2}\right)$$
Now consider $\frac 1z= B_1$ and $1-\frac 1z= B_2$ to get (I think) :
$$-\mathrm{Li}_2\left(\frac 1{B_1}\right)-\mathrm{Li}_2\left(\frac 1{B_2}\right)=\frac{\log\left(\frac {B_2}{B_1}\right)^2}2-\frac{\pi^2}2-i\pi\log\left(B_1 B_2\right)$$
Hoping this helped,
