# Category Archives: Polylogarithm

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$ $$\textit{proof}$$ Note that $$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$ By differentiating with respect to $a$ , $p$ times we have $$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$ $$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$ Let $a =k$ $$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$ Use the relation to polygamma H^{(p)}_k … Continue reading Posted in Euler sum, PolyGamma, Polylogarithm | | Leave a comment ## Generalized nonlinear polylogarithm integral \begin{align*} \int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) \\&-\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align*}\textit{proof}$$We can see that$$\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx = \sum_{k=1}^\infty\sum_{n=1}^\infty\frac{1}{k^{q}n^{p}(n+k)}$$Let us first look at the following$$\mathscr{C}(\alpha , k) =\sum_{n=1}^\infty\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k} $$This can be … Continue reading Posted in Euler sum, Polylogarithm | | Leave a comment ## Nonlinear Euler sums using Nielsen formula According to Nielsen we have the following : If$$f(x)= \sum_{n= 0}^\infty a_n x^n $$Then we have the following$$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\frac{1}{x}\sum_{n=1}^\infty \frac{a_{n-1} H_{n}}{n^2}x^n$$Now let $a_n = H_n$ then we have the … Continue reading Posted in Euler sum, Polylogarithm | Tagged , , , , | Leave a comment ## Special values of the dilgoarithm function Prove that$$\mathrm{Li}_2\left( \frac{\sqrt{5}-1}{2} \right) = \frac{\pi^2}{10} – \log^2 \left( \frac{\sqrt{5}-1}{2}\right)\textit{proof}$$Use the following functional equation$$\mathrm{Li}_2 \left( \frac{z}{z-1} \right) + \frac{1}{2} \mathrm{Li}_2 (z^2) – \mathrm{Li}_2(-z) = -\frac{1}{2} \log^2 (1-z) $$These are proved here and here Now let … Continue reading Posted in Dilogarithm, Polylogarithm | Leave a comment ## Square difference formula for polylgoarithm proof$$\mathrm{Li}_{\,n}(-z) + \mathrm{Li}_{\,n}(z) = 2^{1-n} \,\mathrm{Li}_{\,n}(z^2) \textit{proof}$$As usual we write the series representation of the LHS$$\sum_{k=1}^\infty \frac{z^k}{k^n}+\sum_{k=1}^\infty \frac{(-z)^k}{k^n}$$Listing the first few terms$$z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\cdots +\left(-z+\frac{z^2}{2^n}-\frac{z^3}{3^n}+\cdots \right) $$The odd terms will cancel$$2\frac{z^2}{2^n}+2\frac{z^4}{4^n}+2\frac{z^6}{6^n}+\cdots $$Take $2^{1-n} … Continue reading Posted in Dilogarithm, Polylogarithm | | 1 Comment ## Dilogarithm difference formula proof \mathrm{Li}_2(z) + \mathrm{Li}_2 \left(\frac{z}{z-1} \right) = – \frac{1}{2} \log^2 (1-z) \,\,\,\, \, z<1 \textit{proof} Start by the following \mathrm{Li}_2 \left(\frac{z}{z-1} \right) = -\int^{\frac{z}{z-1}}_0 \frac{ \log(1-t)}{t}\, dt Differentiate both sides with respect to z \frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = \frac{1}{(z-1)^2}\left( \frac{ \log … Continue reading Posted in Dilogarithm, Polylogarithm | Tagged , , , | 1 Comment ## Dilogarithm at 2 \mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{12}-\frac{1}{2}\log^2 \left(\frac{1}{2}\right) proof Using the duplication formula proved here \mathrm{Li}_2\left(z\right)+\mathrm{Li}_2(1-z)\, = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\, We can easily deduce that for \( z=\frac{1}{2}$$$2\mathrm{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{6}-\log^2\left(\frac{1}{2}\right) \,\,$$It follows by dividing by 2. Posted in Dilogarithm, Polylogarithm | Tagged , | Leave a comment ## Dilogarithm functional equation proof$$\mathrm{Li}_2(z) + \mathrm{Li}_{2}(1-z) = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,\,\, ,\,0<z<1\textit{proof}$$Start by the following$$\mathrm{Li}_2\left(z\right) = -\int^{z}_0 \frac{\log(1-t)}{t} \, dt $$Now integrate by parts to obtain$$\mathrm{Li}_2\left(z\right)= -\int^z_0 \frac{\log(t)}{1-t} \, dt -\log(z) \log(1-z)  By the change of variable \(t=1-x … Continue reading

Posted in Dilogarithm, Polylogarithm | | 1 Comment