Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | |||
|
riemann_zeta_function [2016/06/03 22:50] nikolaj |
riemann_zeta_function [2016/06/03 22:54] nikolaj |
||
|---|---|---|---|
| Line 83: | Line 83: | ||
| $\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)$ | $\sum_{n=1}^\infty ({\mathrm e}^{-x})^n = \frac{1} { {\mathrm e}^{-x}-1} = \frac{1} {x} - \frac{1}{2} + \frac{1}{12}x+O(x^2)$ | ||
| + | |||
| + | (check signs) | ||
| The above sum over the integral is convergent for $s>1$, while the expression | 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 $ | $\frac{1} { \Gamma(s) } \int_0^\infty \frac{x^{s-1}} { {\mathrm e}^x-1} \, {\mathrm d}x $ | ||
| + | |||
| + | (check signs) | ||
| works for all complex $s\ne 1$. We thus found the analytical continuation. | works for all complex $s\ne 1$. We thus found the analytical continuation. | ||
