Monthly Archives: January 2017

Pfaff transformations of the hypergeometric function proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;\frac{z}{z-1}\right)$$ and $${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;\frac{z}{z-1}\right)$$ $$\textit{proof}$$ Start by the integral representation $${}_2F_1 \left(a,b;c;z\right)=\frac{1}{\beta(b,c-b)}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$ By the the transformation \( t\to 1-t \) $$\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-(1-t)z)^a}\, dt=\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-z+tz)^a}\, dt$$ Which can be written as $$\frac{(1-z)^{-a}}{\beta(b,c-b)}\int_0^1t^{c-b-1}(1-t)^{b-1} \left( 1-t\frac{z}{z-1} … Continue reading

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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 $$i(f(ix) -f(-ix))= \frac{i}{z+ix}-\frac{i}{z-ix} = \frac{2x}{z^2+x^2}$$ 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

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Sum of natural numbers equal -1/12 ?

$$\tag{1}1+2+3+4+\cdots =^?-\frac{1}{12}$$ There are many problems with that equality. Can a summation of positive integers lead to a negative quantity ? The equality implies that $$\sum_{k=1}^\infty k = -\frac{1}{12}$$ But we already know that $$S_N = \sum_{k=1}^Nk = \frac{N(N+1)}{2} \to … Continue reading

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Integral representation of Gauss Hypergeometric function proof

$$\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt $$ $$\textit{proof}$$ Start by the RHS $$\int_0^1 t^{b-1}(1-t)^{c-b-1} \, (1-tz)^{-a}\, dt $$ Using the expansion of \( (1-tz)^{-a} \) we have $$\int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{k=0}^\infty\frac{(a)_k}{k!}\, (tz)^k $$ Interchanging the integral with the series $$ \sum_{k=0}^\infty\frac{(a)_k}{k!}\, z^k … 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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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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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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Stirling numbers of first kind generating function

Prove the following $$\sum_{n=k}^\infty(-1)^{n-k}\left[n\atop k\right] \frac{z^n}{n!} = \frac{\log^k(1+z)}{k!}$$ $$\textit{proof}$$ We start by the following $$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n$$ Now use that $${u \choose n} = \frac{\Gamma(u+1)}{\Gamma(u-n+1)n!}$$ Now use that $$\frac{\Gamma(u+1)}{\Gamma(u-n+1)} = \frac{u(u-1)\cdots (u-n+1)\Gamma(u+1)}{\Gamma(u+1)} = (u)_n$$ This implies … Continue reading

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