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

## 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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## 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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## 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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## 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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## Signed Stirling numbers of first kind as coefficients

Signed Stirling numbers of the first kind We define the following $$s(n,k) = (-1)^{n-k} \left[n\atop k\right]$$ Prove the following $$(x)_n = x(x-1)(x-2)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k$$ $$\textit{proof}$$ We already proved that $$x^{(n)} = \sum_{k=0}^n \left[n\atop k\right] x^k$$ Which can be … 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

Posted in Dilogarithm, Polylogarithm | | 1 Comment

$$\mathrm{Li}_2(z) + \mathrm{Li}_2 \left(\frac{z}{z-1} \right) = – \frac{1}{2} \log^2 (1-z) \,\,\,\, \, z<1$$ $$\textit{proof}$$ 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 $z$ $$\frac{d}{dz}\mathrm{Li}_2 \left(\frac{z}{z-1} \right) = \frac{1}{(z-1)^2}\left( \frac{ \log … Continue reading Posted in Dilogarithm, Polylogarithm | Tagged , , , | 1 Comment ## Dilogarithm functional equation proof$$\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

Posted in Dilogarithm, Polylogarithm | | 1 Comment