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Zeta September 8, 2007

Filed under: Analysis,Funktion,Primzahlen,Reihen,Zahlentheorie — abi007 @ 7:23 am

\displaystyle\zeta(n)=\sum_{k=1}^{\infty}\frac1{k^n}
\displaystyle\zeta(s) = \prod_{p\ \mathrm{prim}}\frac1{1-1/p^s}=\frac1{(1-1/2^s)(1-1/3^s)(1-1/5^s)\cdots}.
\displaystyle \zeta_K(s)=\sum_{I\subset\mathcal{O}_K} (N^K_{\mathbb{Q}}(I))^{-s}= \prod_{\mathfrak{p}}\frac{1}{1-(N^K_{\mathbb{Q}}(\mathfrak{p}))^{-s}}

\displaystyle\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}

\displaystyle \zeta(2)=\frac{\pi^2}6

http://www.mathlinks.ro/viewtopic.php?t=159424&start=20

http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

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\displaystyle\gamma:=\lim_{n\to\infty} \left(H_n-\ln n\right)= \lim_{n\rightarrow \infty}\; \sum_{i=1}^n \left[\frac{1}{i} - \ln \left( 1 + \frac{1}{i} \right) \right]

\displaystyle\gamma= -\int\limits_0^1 \ln(-\ln x)\, \mathrm{d}x

\displaystyle\gamma = -\int\limits_0^\infty e^{-x}\ln x\, \mathrm{d}x

\displaystyle\gamma = -\int\limits_0^\infty (\frac1{1-e^{-x}}-\frac1x)e^{-x} \mathrm{d}x

\displaystyle\gamma = -\int\limits_0^\infty \frac1x(\frac1{1+x}-e^{-x}) \mathrm{d}x

\displaystyle \gamma\;\;=\;\;\int_{0}^{1}\;\;\frac{1}{x+1}\cdot\left(\sum_{k=1}^{\infty}\;x^{2^{k}-1}\right)\;\textbf dx

\displaystyle \gamma\;\;=\;\; \int_{0}^{1}\;\;\frac{x+2}{x^{2}+x+1}\cdot\left(\sum_{k=1}^{\infty}\;x^{3^{k}-1}\right)\;\textbf dx

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\displaystyle \gamma = - \Gamma'(1)

\displaystyle \Gamma(z) =\lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)}

\displaystyle\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}(1 + \frac{z}{n})^{-1}e^{\frac{z}{n}}

\displaystyle \Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt

\displaystyle =[-t^{x-1}e^{-t}]_0^\infty+\int_0^\infty(x-1)t^{x-2}e^{-t}dt

\displaystyle =(x-1)\int_0^\infty t^{x-2}e^{-t}dt

\displaystyle =(x-1)\Gamma(x-1).

\displaystyle\Gamma \left( x \right)\Gamma \left( {1 - x} \right) = \pi \csc \pi x

\displaystyle\int_{0}^{\infty} \frac{x^n}{e^x-1}=\zeta(n+1)\Gamma(n+1)

http://en.wikipedia.org/wiki/Planck_law#Appendix

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\displaystyle\int_0^{\frac{1}{2}} \; \;\frac{{\tan ^{ - 1} \,x}}{x}\;\;{\mathbf{d}}x = \sum\limits_{n = 0}^{ + \infty } {\frac{{( - 1)^n }}{{(2n + 1)^2 \cdot2^{2n + 1} }}}

\displaystyle{\rm K} = \beta (2) = \sum\limits_{n = 0}^\infty {\frac{{( - 1)^n }}{{(2n + 1)^2 }}} = \frac{1}{{1^2 }} - \frac{1}{{3^2 }} + \frac{1}{{5^2 }} - \frac{1}{{7^2 }} + \cdots

\displaystyle = - \int_0^1 {\frac{{\ln (t)}}{{1 + t^2 }}} {\text{ d}}t

\displaystyle = \int_0^1 {\int_0^1 {\frac{1}{{1 + x^2 y^2 }}} } dxdy = \frac{1}{2}\int_0^1 {\text{K}} (x)\,dx = \int_0^1 {\frac{{\tan ^{ - 1} x}}{x}} dx

 

 
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