Consider the following limit: $$ Z(\beta) = \lim_{z\to1-}\int_0^1 \frac{v^\beta\,dv}{1-z v}\log\frac{1-v z}{1-v}. $$ (This is related to this question.)

What is the closed form for this limit?

Numerically, I suspect that $$ Z(\beta) = \frac{\pi^2}{6}, $$ independently of $\beta$, but I have been unable to prove this.


1 Answer 1


That the limit is independent of $\beta$ is rather clear. Consider the difference of two such integrals: $$ I(\beta_2)-I(\beta_1)=\int_0^1\frac{v^{\beta_2}-v^{\beta_1}}{1-zv}\ln\frac{1-zv}{1-v}\,dv\tag{1} $$ The prefactor $\displaystyle\frac{v^{\beta_2}-v^{\beta_1}}{1-zv}$ is no longer singular as we set $z=1$, $v\rightarrow1$. Therefore, to compute the limit of (1) as $z\rightarrow1$, we can simply set $z=1$ inside, which gives $Z(\beta_2)-Z(\beta_1)=0$.

Now it suffices to compute the integral for some arbitrary value of $\beta$, for example, for $\beta=0$. But such integral can be expressed in terms of dilogarithms: $$\int_0^1\frac{1}{1-vz}\ln\frac{1-vz}{1-v}\,dv=\frac{\mathrm{Li}_2(z)}{z}.$$ The limit of the last expression as $z\rightarrow 1$ is $\mathrm{Li}_2(1)=\zeta(2)=\frac{\pi^2}{6}$.

  • $\begingroup$ Great answer, very clear. Thanks. $\endgroup$
    – Kirill
    Jun 8, 2013 at 15:41

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