Category Archives: Polylogarithm

Integral representation of generalized Euler sums

$$\sum_{k=1}^\infty\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$ $$\textit{proof}$$ 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

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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

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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

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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

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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

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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

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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.

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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

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