#dilogarithm

[thetwomeatmeal]

Until today, I suffered from the following misconception:

Misconception. A complex function that is analytic in a neighborhood some point has a singularity on the boundary of its disk of convergence.

Below the jump, I present a counterexample, give a correct statement, and say why I care.

The counterexample is the dilogarithm function, $$\mathrm{Li}_2(z) = \sum_{n=0}^\infty \frac{z^n}{n^2}.$$ So named because of its similarity to a function I hope you won’t mind if I call “the logarithm”, $$\mathrm{Li}_1(z) = -\log(1-z) = \sum_{n=0}^\infty \frac{z^n}{n}.$$ By the way, observe that $$\frac{d}{dz}\mathrm{Li}_2(z) = \frac{-\log(1-z)}{z}.$$

There’s a cool formula, due to Hadamard, that the radius of convergence of $\sum a_n z^n$ is $(\lim\sup |a_n|^{1/n})^{-1}$. (I forgot the reciprocal in a previous version of this post.) This gives radius of convergence 1 for both the logarithm and the dilogarithm. The function $\mathrm{Li}_1(z)$ has a pretty nasty singularity at $z = 1$, and the series diverges there, too -- it’s the harmonic series. On the other hand, $\mathrm{Li}_2(1) = \pi^2/6$, and so the dilogarithm converges (absolutely!) everywhere on the unit circle. What’s going on?

Let’s look closer at the singularity of $\mathrm{Li}_1$ at 1. We can analytically continue the function around this singularity by requiring that $\mathrm{Li}_1′(z) = \frac{1}{1-z}$. However, if we take a point near 1 and move it (counterclockwise) in a circle around 1, we add a factor of $$\oint_{|z| - 1 = \epsilon} \frac{1}{1-z}\,dz = -2\pi i$$ to its logarithm. So this singularity is a branch point -- the logarithm does not extend to a meromorphic function in any neighborhood of it, and to even get a well-defined extension to most of the complex plane, we need to make it impossible to go in a circle around the branch point. Typically, this is done by removing the ray $[1,\infty)$.

Now, if the dilogarithm extended to an analytic function in any neighborhood of 1, its derivative would be an analytic extension of $\mathrm{Li}_1(z)/z$, and so the logarithm would be analytic in this neighborhood! And again, we can make an analytic extension of $\mathrm{Li}_2$ to $\mathbb{C} - [1,\infty)$, by defining $$\mathrm{Li}_2(z) = \int_0^z \mathrm{Li}_1(z)\, dz,$$ where the integration is along a path that doesn’t cross $[1,\infty)$.

Another take on what’s going on is that the dilogarithm extends continuously but not differentiably to this boundary point. In fact, the “one-sided derivatives” of the dilogarithm along paths that go to 1 from inside the disk go off to $\mathrm{Li}_1(1) = \infty$.

Likewise, we can define $$\mathrm{Li}_m(z) = \sum_{n=0}^\infty \frac{z^n}{n^m}$$. This extends analytically to $\mathbb{C} - [1,\infty)$ and is exactly “m-1 times differentiable” at the boundary point 1. Is there a power series, though, that:

converges on the open disk of radius 1,

extends continuously to the closed disk of radius 1,

doesn’t extend analytically to any open neighborhood of 1,

does extend analytically to an open neighborhood of every other point on the boundary,

and such that all of its derivatives also have these properties?

Why I care: I mean, I don’t care a hoot about complex functions, but I’ve been doing things with $p$-adic power series that have required me to revisit these basic analysis concepts. I had a power series that converged in a small $p$-adic disk, which I was trying to turn into one that converged everywhere on $\mathbb{Z}_p$, and I’d been thinking that it had singularities on the boundary on the disk. This is no more true in the $p$-adic world than in the Archimedean one. For example, the power series $$f_1(x) = \sum_{n=0}^\infty p^nx^{p^n}, \quad f_2(x) = \sum_{n=0}^\infty p^{-n} x^{p^n}$$ both have radius of convergence 1 (by the same Hadamard criterion), but the first one converges everywhere on the boundary of the unit disk $\mathbb{Z}_p$, the second one nowhere! (I don’t think I’m screwed, though -- the lesson might just be that “radius of convergence” is a more basic concept than “singularity” in this context, and I think I can still work with it.) 

[mathematicianadda]

How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?

Can we evaluate $\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$ ?

where $H_n=\sum_{k=1}^n\frac1n$ is the harmonic number.

A related integral is $\displaystyle\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2\left(\frac x2\right)}{x}dx$.

where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$ is the dilogarithmic function.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/34bPfkH

[mathematicianadda]

Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$

I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$

My Attempt:

I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$ where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$

Evaluating this integral by integration by parts and geometric series I get $$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$ Re-indexing I get $$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$ $$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at $$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ $$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$

At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger http://bit.ly/2HMuO4j