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riemann_zeta_function [2016/06/02 10:58] nikolaj |
riemann_zeta_function [2016/06/03 22:54] (current) nikolaj |
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| ${\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 | + | (check signs) |
| + | |||
| + | 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 $ | ||
| + | |||
| + | (check signs) | ||
| - | $\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 the $ \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 | ||
