LOG#162. Polylogia flashes(IV).

polylog

In this final post (by the moment) in the polylogia series we will write some additional formulae for polylogs and associated series.

Firstly, \forall s\in C\neq 0,1,2,3 we have

(1)   \begin{equation*} Li_s(e^\mu)=\dfrac{\Gamma (1-s)}{(2\pi)^{1-s}}\left[i^{1-s}\zeta (1-s,\dfrac{\mu}{2\pi i})+i^{s-1}\zeta (1-s,1-\dfrac{\mu}{2\pi i})\right] \end{equation*}

and now, if 0<Im(\mu)\leq 2\pi

(2)   \begin{equation*} Li_s(e^\mu)=\Gamma(1-s)\left[(-2\pi i)^{s-1}\sum_{k=0}^\infty\left(k+\dfrac{\mu}{2\pi i}\right)^{s-1}+(2\pi i)^s\sum_{k=0}^\infty\left(k+1-\dfrac{\mu}{2\pi i}\right)^{s-1}\right] \end{equation*}

(3)   \begin{equation*} Li_s(e^\mu)=\Gamma (1-s)\sum_{k=-\infty}^\infty\left(2\pi i k-\mu\right)^{s-1} \end{equation*}

\forall Re(s)<0,\forall \mu/e^\mu\neq 1

The next identity also holds

    \[Li_s(e^\mu)=\Gamma (1-s)(-\mu)^{s-1}+\sum_{k=0}^\infty\dfrac{\zeta (s-k)\mu^k}{k!}\]

The Bose-Einstein integral can be rewritten for Re(s)>0 as follows

    \[Li_s(z)=\dfrac{z}{2}+\dfrac{z}{2\Gamma (s)}\int_0^\infty e^{-t}t^{s-1}\coth\left(\dfrac{t-\ln z}{2}\right)dt\]

and

    \[Li_s(z)=\dfrac{z}{2}+\sum_{k=-\infty}^\infty\dfrac{\Gamma (1-s,2\pi ik-\ln z)}{(2\pi ik-\ln z)^{1-s}}\]

The polylog has the following asymptotic series. If \vert z\vert >>1 we have

(4)   \begin{equation*} Li_s(z)=\dfrac{\pm iz}{\Gamma (s)}\left[\ln (-z)\pm i\pi\right]^{s-1}-\sum_{k=0}^\infty(-1)^k(2\pi)^{2k}\dfrac{B_{2k}}{(2k)!}\dfrac{\left[\ln(-z)\pm i\pi\right]^{s-2k}}{\Gamma (s+1-2k)} \end{equation*}

(5)   \begin{equation*} Li_s(z)=\sum_{k=0}^\infty(-1)^k(1-2^{1-2k})(2\pi)^{2k}\dfrac{B_{2k}}{(2k)!}\dfrac{\left[\ln(-z)\right]^{s-2k}}{\Gamma (s+1-2k)} \end{equation*}

Euler’s dilogarithm or Spence’s function has nice features. Interestingly, computer algebra systems generally define the dilogarithm as Li_2(1-z), thus be aware with the definitions you read, use and write. In our case:

(6)   \begin{equation*} Li_2(z)=-\int_0^z\dfrac{\ln (1-t)}{t}dt=-\int_0^1\dfrac{\ln (1-zt)}{t}dt \end{equation*}

Moreover, for Re(z)\geq 1 you can find

(7)   \begin{equation*} Li_2(z)=\dfrac{\pi^2}{6}-\int_1^z\dfrac{\ln (t-1)}{t}dt-i\pi \ln (z) \end{equation*}

(8)   \begin{equation*} Li_2(z)=\dfrac{\pi^2}{3}-\dfrac{1}{2}\left(\ln (z)\right)^2-\sum_{k=1}^\infty\dfrac{1}{k^2z^k}-i\pi\ln z \end{equation*}

\forall x\notin (1,\infty) and \notin (1,\infty) we have the Abel identity:

(9)   \begin{equation*} \ln(1-x)\ln(1-y)=Li_2\left(\dfrac{x}{1-y}\right)+Li_2\left(\dfrac{y}{1-x}\right)-Li_2(x)-Li_2(y)-Li_2\left(\dfrac{xy}{(1-x)(1-y)}\right) \end{equation*}

Euler’s reflection formula with y=1-x follows up

(10)   \begin{equation*} Li_2(x)+Li_2(1-x)=\dfrac{\pi^2}{6}-\ln (x)\ln(1-x) \end{equation*}

    \[Li_2(1)=\zeta (2)=\dfrac{\pi^2}{6}\]

The pentagon identity, with u=\dfrac{x}{1-y} and v=\dfrac{y}{1-x} is a mutation of the Abel identity

    \[\boxed{Li_2(u)+Li_2(v)-Li_2(uv)=Li_2\left(\dfrac{u-uv}{1-uv}\right)+Li_2\left(\dfrac{v-uv}{1-uv}\right)+\ln\left(\dfrac{1-u}{1-uv}\right)\ln\left(\dfrac{1-v}{1-uv}\right)}\]

Landen’s identity is also beautiful. It arises if we write y=x in the Abel identity and we square the relationship we get:

    \[Li_2(-x)=-Li_2\left(\dfrac{x}{1+x}\right)-\dfrac{1}{2}\left[\ln\left(x+1\right)\right]^2\]

with x\notin (-\infty,-1). We have an inversion formula for this too

    \[Li_2(x)+Li_2\left(\dfrac{1}{x}\right)=-\dfrac{\pi^2}{6}-\dfrac{1}{2}\left(\ln(-x)\right)^2\]

with x\notin (0,1), or

    \[Li_2(x)+Li_2\left(\dfrac{1}{x}\right)=\dfrac{\pi^2}{3}-\dfrac{1}{2}\left(\ln x\right)^2-i\pi\ln x\]

with x\geq 1.

The mathematician Don Zagier has stated the following sentence “(…)The dilogarithm is the only mathematical functionwith a sense of humor(…)”. Have a look at the following values of Euler’s dilogarithm:

    \[Li_2(-1)=-\dfrac{\pi^2}{12}\]

    \[Li_2(0)=0\]

    \[Li_2(1/2)=\dfrac{\pi^2}{12}-\dfrac{\ln^22}{2}\]

    \[Li_2(1)=\dfrac{\pi^2}{6}=\zeta(2)\]

    \[Li_2(2)=\dfrac{\pi^2}{4}-i\pi\ln 2\]

Moreover, suppose that x=-\phi=-\dfrac{1+\sqrt{5}}{2} is the negative golden mean ratio, then

    \[Li_2(-\phi)=-\dfrac{\pi^2}{10}-\ln^2\left(\dfrac{\sqrt{5}-1}{2}\right)\]

If x=\phi-1=1/\phi we also have

    \[Li_2(1/\phi)=Li_2\left(\phi-1\right)=-\dfrac{\pi^2}{15}+\dfrac{1}{2}\ln^2\left(\dfrac{\sqrt{5}-1}{2}\right)\]

If x=-\dfrac{1}{2}\left(3-\sqrt{5}\right)\equiv\Xi, we have

    \[Li_2(\Xi)=\dfrac{\pi^2}{15}-\ln^2\left(\dfrac{\sqrt{5}-1}{2}\right)\]

and

    \[Li_2(1-\phi)=\dfrac{\pi^2}{10}-\ln^2\left(\dfrac{\sqrt{5}-1}{2}\right)\]

    \[Li_2(\phi)=\dfrac{11\pi^2}{15}+\dfrac{1}{2}\ln^2\left(-\left(\dfrac{\sqrt{5}-1}{2}\right)\right)\]

    \[Li_2\left(\dfrac{\sqrt{5}+3}{2}\right)=-\dfrac{11\pi^2}{15}-\ln^2\left(-\left(\dfrac{\sqrt{5}+1}{2}\right)\right)\]

 There are also polylog ladders much more complex than the above identities. Let us define

    \[\rho=\dfrac{1}{2}\left(\sqrt{5}-1\right)=\phi-1=\dfrac{1}{\phi}\]

Then, a wonderful result by Coxeter (1935) is the next identity

(11)   \begin{equation*} Li_2(\rho^6)=4Li_2(\rho^3)+3Li_2(\rho^2)-6Li_2(\rho)+\dfrac{7}{30}\pi^2 \end{equation*}

and Landen also derived

(12)   \begin{equation*} Li_2(\rho)=\dfrac{\pi^2}{10}-\ln^2(\rho) \end{equation*}

Now, we can write some multiplication theorems. The duplication identity

(13)   \begin{equation*} 2^{1-s}Li_s(z^2)=Li_s(z)+Li_s(-z) \end{equation*}

Gauss wrote the next sum, a discrete Fourier transform

(14)   \begin{equation*} k^{1-s}Li_s(z^k)=\sum_{n=0}^{k-1}Li_s\left(ze^{2\pi in/k}\right) \end{equation*}

The Kummer’s identity for duplication reads

(15)   \begin{equation*} 2^{1-n}\Lambda_n(-z^2)=\Lambda_n(z)+\Lambda_n(-z) \end{equation*}

Moreover, if Char F=0, we can derive the result

    \[\lambda^{-\nu}J_\nu(\lambda z)=\sum_{n=0}^\infty\dfrac{1}{n!}\left(\dfrac{(1-\lambda^2)z}{2}\right)^nJ_{\nu+n}(z)\]

and where J_{\nu}(z) is the Bessel function with \lambda,\nu\in C.

Periodic zeta functions are defined by

(16)   \begin{equation*} F(s,q)=\sum_{m=1}^\infty\dfrac{e^{2\pi imq}}{m^s}=Li_s\left(e^{2 \pi i q}\right) \end{equation*}

and thus

    \[2^{-s}F(s,q)-F(s,\frac{q}{2})+F(s,\frac{q+1}{2})\]

    \[k^{-s}F(s,kq)=\sum_{n=0}^{k-1}F(s,q+\frac{n}{k})\]

Finally, we end this series with some Hurwitz zeta function series and identities

    \[k^s\zeta(s)=\sum_{n=1}^k\zeta(s,\frac{n}{k})\]

    \[k^s\zeta(s,kz)=\sum_{n=0}^{k-1}\zeta (s,z+\frac{n}{k})\]

    \[\zeta(s,kz)=\sum_{n=0}^\infty\begin{pmatrix}s+n-1\\ n\end{pmatrix}(1-k)^nz^n\zeta(s+n,z)\]

See you in my next blog post!

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