Tag Archives: Euler

General formula for an integral involving powers of logarithms

$$\int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l’}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j}$$ $$\textit{solution}$$ Stirling numbers of the first kind might be useful here, Consider $$m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{x^k}{k!} = \log^m(1+x)$$ $$\int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx = … Continue reading

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Multiple integral related to Euler sums

Posted by Cornel Ioan Valean on Facebook Show that $$\small \int^1_0 \cdots \int^1_0 \frac{\mathrm{d}\varphi_1\cdots \mathrm{d}\varphi_5}{(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)} = \frac{59}{32}\zeta(5)-\frac{1}{2}\zeta(2)\zeta(3)$$ $$\textit{proof}$$ Consider the integral $$I = \int^1_0 \cdots \int^1_0 \frac{\mathrm{d}\varphi_1\cdots \mathrm{d}\varphi_5}{(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)(\varphi_1+\varphi_2+\varphi_3+\varphi_4+\varphi_5+1)}$$ Use the series expansion $$I = \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{(-1)^{n+k}}{n+1} \int^1_0\cdots \int^1_0 (\varphi_1\cdots … Continue reading

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Solving Euler sums using Contour integration

Prove that $$\sum_{n=1}^\infty \frac{H_n}{n^2} = 2\zeta(3)$$ $$\textit{proof}$$ Consider the function $$f(z) = \frac{(\psi(-z)+\gamma)^2}{z^2}$$ Note that \( f \) has poles at non-negative integers By integration around a large circle \( |z| = \rho \) Note that $$\oint f(z)\,dz = 2\pi … Continue reading

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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 multi-valued defined as $$\log(z) = \ln|z|+i\theta +2k\pi i$$ This maps the complex plain more than once … Continue reading

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Euler Hypergeometric transformation proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{c-a-b} {}_2F_1 \left(c-a,c-b;c;z\right)$$ $$\textit{proof}$$ In the Pfaff transformations let \( z \to \frac{z}{z-1}\) , proved here  $${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)$$ and $${}_2F_1 \left(a,b;c;\frac{z}{z-1}\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$ By equating the two transformations $$(1-z)^{-a} {}_2F_1 \left(a,c-b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;z\right)$$ Now use the transformation … Continue reading

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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 $$\sum_{i=1}^\infty \,\sum_{k=i}^\infty\frac{1}{i^p} \frac{1}{k^q}=\sum_{i=1}^\infty \,\sum_{k=1}^\infty\frac{1}{i^p} \frac{1}{k^q}-\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q}$$ The second sum can be written as $$\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q} = \sum_{i=1}^\infty\frac{1}{i^p} \,\left(\sum_{k=1}^{i} \frac{1}{k^q}-\frac{1}{i^p}\right)$$ By separating and … Continue reading

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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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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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Nonlinear euler sum proof using stirling numbers of the first kind

Prove that $$\sum_{k=1}^\infty \frac{(H_k)^2}{k^2} = \frac{17\pi^4}{360}$$ $$\textit{proof}$$ Start by the following which can be proved by induction $$\frac{\left[n\atop 3\right]}{n!} =\frac{ (H_{n-1})^2-H^{(2)}_{n-1}}{2n}$$ And the generating function proved here $$-\sum_{n=3}^\infty \left[n\atop 3\right] \frac{z^n}{n!} = \frac{\log^3(1-z)}{6}$$ Hence we get $$\sum_{n=3}^\infty ( H^{(2)}_{n-1}- (H_{n-1})^2) … Continue reading

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Gamma Euler Reflection formula proof

$$\Gamma \left({z}\right) \Gamma \left({1 – z}\right) = \frac \pi {\sin \left({\pi z}\right)}\,\,\, \, , \forall z \notin \mathbb{Z}$$ $$\textit{proof}$$ We can use the sine product formula $$\frac{\pi}{\sin(\pi z)} = \frac{1}{z}\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)^{-1}$$ Now we start by noting that $$\Gamma(z)\Gamma(1-z) = -z\Gamma(z)\Gamma(-z) $$ … Continue reading

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