Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Last revision Both sides next revision | ||
riemann_zeta_function [2016/05/31 20:20] nikolaj |
riemann_zeta_function [2016/06/03 22:50] nikolaj |
||
---|---|---|---|
Line 20: | Line 20: | ||
== Functional equation == | == Functional equation == | ||
Tells you most values: | Tells you most values: | ||
- | ^ $ \zeta(s) = 2(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s)$ ^ | + | ^ $ \zeta(s) = 2\,(2\pi)^{s-1}\sin{\left(\pi\,s/2\right)}\,\Gamma(1-s)\,\zeta(1-s)$ ^ |
== Specific values == | == Specific values == | ||
Line 27: | Line 27: | ||
so that | so that | ||
+ | |||
+ | $\zeta(1-2m) = \dfrac{2(2m-1)!}{(4\pi^2)^m}\cos(m\pi)\zeta(2m)$ | ||
$\zeta(1-2m)=(-1)^{m+1}\frac{1}{2m}B_{2m}$ | $\zeta(1-2m)=(-1)^{m+1}\frac{1}{2m}B_{2m}$ | ||
Line 74: | Line 76: | ||
${\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x $ | ${\frac {1} {n^s}} = {\frac {1} {\Gamma(s)}} \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x $ | ||
- | and then, recognizing the geometric series | + | and thus |
+ | |||
+ | $\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \sum_{n=1}^\infty \int_0^\infty x^{s-1} {\mathrm e}^{-nx}\,{\mathrm d}x $ | ||
+ | |||
+ | Now consider the geometric series | ||
$\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)$ | $\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)$ | ||
- | give you | + | The above sum over the integral is convergent for $s>1$, while the expression |
+ | |||
+ | $\frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x $ | ||
- | $\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x $ | + | works for all complex $s\ne 1$. We thus found the analytical continuation. |
- | He takes the integral into the complex plane, where he $ \frac{1} { {\mathrm e}^x-1}$ diverges periodically in steps of $2\pi\,i$. | + | The integrand $ \frac{1} { {\mathrm e}^x-1}$ diverges periodically in steps of $2\pi\,i$. |
He discovers that the function obeys a reflection formula | He discovers that the function obeys a reflection formula | ||