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Tag Archives: formula
Euler reflection formula proof using contour integration
$$\int_{0}^{\infty}\frac{x^{\alpha}}{x+1}\,dx= \pi \csc(\pi \alpha)$$ $$\textit{proof}$$ Consider the following function $$f(z) = \frac{z^{\alpha}}{1+z} = \frac{e^{\alpha \log(z)}}{1+z}$$ As we know the function \( \log(z) \) is multivalued defined as $$\log(z) = \lnz+i\theta +2k\pi i$$ This maps the complex plain more than once … Continue reading
Posted in Beta function, Contour Integration, Gamma function
Tagged analysis, complex, contour, Euler, formula, proof, reflection
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Integral representation of the digamma function using Abel–Plana formula
$$ \int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}1}\,dx =\log(z)\psi(z)\frac{1}{2z}$$ $$\textit{proof}$$ Use Abel–Plana formula $$\sum_{n=0}^\infty f(n) = \int^\infty_0f(x)\,dx+\frac{f(0)}{2} +i\int^\infty_0 \frac{f(ix)f(ix)}{e^{2\pi x}1}\,dx$$ Let $$f(x) = \frac{1}{z+x}$$ Note that $$i(f(ix) f(ix))= \frac{i}{z+ix}\frac{i}{zix} = \frac{2x}{z^2+x^2}$$ By integration we have $$ \int^\infty_0 \frac{2x}{(x^2+z^2)e^{2\pi x}1}\,dx =\lim_{N\to \infty}\sum_{n=0}^N \frac{1}{z+n}\int^N_0 \frac{1}{x+z}\,dx\frac{1}{2z}$$ The … Continue reading
Posted in Digamma function
Tagged Abel–Plana, Digamma, formula, function, Integral, proof, representation
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Symmetry formula for Generalized Linear Euler sums
$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$ $$\textit{proof}$$ Take the leftmost series and swap the finite and infinite sums $$\sum_{i=1}^\infty \,\sum_{k=i}^\infty\frac{1}{i^p} \frac{1}{k^q}=\sum_{i=1}^\infty \,\sum_{k=1}^\infty\frac{1}{i^p} \frac{1}{k^q}\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i1} \frac{1}{k^q}$$ The second sum can be written as $$\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i1} \frac{1}{k^q} = \sum_{i=1}^\infty\frac{1}{i^p} \,\left(\sum_{k=1}^{i} \frac{1}{k^q}\frac{1}{i^p}\right)$$ By separating and … Continue reading
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 $$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt \frac{1}{x}\sum_{n=1}^\infty \frac{a_{n1} H_{n}}{n^2}x^n$$ Now let \( a_n = H_n \) then we have the … Continue reading
Square difference formula for polylgoarithm proof
$$\mathrm{Li}_{\,n}(z) + \mathrm{Li}_{\,n}(z) = 2^{1n} \,\mathrm{Li}_{\,n}(z^2) $$ $$\textit{proof}$$ 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^{1n} … Continue reading
Posted in Dilogarithm, Polylogarithm
Tagged difference, dilogarithm, formula, polylgoarithm, proof, recurrence, square
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Dilogarithm difference formula proof
$$\mathrm{Li}_2(z) + \mathrm{Li}_2 \left(\frac{z}{z1} \right) = – \frac{1}{2} \log^2 (1z) \,\,\,\, \, z<1$$ $$ \textit{proof} $$ Start by the following $$\mathrm{Li}_2 \left(\frac{z}{z1} \right) = \int^{\frac{z}{z1}}_0 \frac{ \log(1t)}{t}\, dt$$ Differentiate both sides with respect to $z$ $$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z1} \right) = \frac{1}{(z1)^2}\left( \frac{ \log … Continue reading
Dilogarithm functional equation proof
$$\mathrm{Li}_2(z) + \mathrm{Li}_{2}(1z) = \frac{\pi^2}{6}\log(z) \log(1z) \,\,\,\, ,\,0<z<1$$ $$\textit{proof}$$ Start by the following $$\mathrm{Li}_2\left(z\right) = \int^{z}_0 \frac{\log(1t)}{t} \, dt $$ Now integrate by parts to obtain $$\mathrm{Li}_2\left(z\right)= \int^z_0 \frac{\log(t)}{1t} \, dt \log(z) \log(1z) $$ By the change of variable \(t=1x … Continue reading
Posted in Dilogarithm, Polylogarithm
Tagged difference, dilogarithm, formula, functional, proof
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