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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.

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2 Answers

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.

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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 :

real partenter image description here

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,

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