# Tag Archives: Euler

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 $$\oint f(z)\,dz = 2\pi … Continue reading Posted in Contour Integration, Euler sum | | Leave a comment ## Euler reflection formula proof using contour integration$$\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$$This maps the complex plain more than once … Continue reading | | Leave a comment ## Euler Hypergeometric transformation proof$${}_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)$$Now use the transformation … Continue reading Posted in Hypergeoemtric function | | Leave a comment ## Symmetry formula for Generalized Linear Euler sums$$\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)$$By separating and … Continue reading Posted in Euler sum | Tagged , , , , , | Leave a comment ## Integral representation of generalized Euler sums$$\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

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