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I am considering the contour integral: $\int Li_2\left( \frac{1-z}{2}\right)Li_2\left( \frac{z-1}{2z}\right) \frac{dz}{z}$.

The contour of integration is the unit circle excluding the pole $z = 0$. $Li_2(z)$ is the dilogarithm function also known as Spence function. We thus have four parts to evaluate in this integration:

a - The unit circle in which we set $z = e^{it}$, $0<t<2 \pi$

b - The real negatively oriented segment, set $z = x$, $x$ runs from $1$ to $0$

c - The small limit integral , set $z = \epsilon e^{it}$, $t$ goes from $2 \pi$ to $0$.

d - The real positively oriented segment, set $z = x$, $x$ runs from $0$ to $1$

It is obvious integrals in (b) and (d) cancel each other, and (a) is easy to set and it is finite. However, I am having a problem in solving integral (c), when you set $\epsilon \rightarrow 0$, the integral diverges having the term $\ln^2(\epsilon)$. I am using the relation $Li_2(z) + Li_2(\frac{1}{z}) = -\frac{\pi^2}{6} - \frac{1}{2} \ln^2(-z)$ for small $z$

Applying Cauchy Integral theorem, this contour integral should be zero because the function is analytic everywhere in the open unit disk except at $z = 0$ and we are excluding this point in our contour. Thus $I_a = -I_c$.

Any hint on how to evaluate $I_c$?

Thank you for your valuable support

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  • $\begingroup$ I don't understand what is your contour. $\text{Li}_2(\frac{1-1/z}{2})$ isn't meromorphic on $|z| \le 1$, it has a branch point at $z=0$. But $\text{Li}_2(\frac{1-z}{2})(\text{Li}_2(\frac{2z}{z-1})-\frac{\pi^2}{6})$ is analytic for $|z|\le r$ small enough. $\endgroup$
    – reuns
    Commented Aug 29, 2017 at 12:28
  • $\begingroup$ Hi, My contour is visualized on this site: www-thphys.physics.ox.ac.uk/people/FrancescoHautmann/… $\endgroup$
    – Hmath
    Commented Aug 29, 2017 at 13:02
  • $\begingroup$ I am using a branch cut from 0 to 1 in order to exclude the singularity z = 0. The function becomes analytic inside this contour. I did not understand why you subtracted $\frac{\pi^2}{6}$ from the second term of the function. $\endgroup$
    – Hmath
    Commented Aug 29, 2017 at 13:04
  • $\begingroup$ In that case $\int_C f(z)dz = 0$ as $f$ is analytic inside $C$. Next time copy and paste the contour as an image $\endgroup$
    – reuns
    Commented Aug 29, 2017 at 13:05
  • $\begingroup$ I definitely know it is 0 by Cauchy integral theorem. My problem is that $I_a$ is finite while $I_c$ is infinite. thus their sum cannot be zero!! $\endgroup$
    – Hmath
    Commented Aug 29, 2017 at 13:06

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