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galoisjr
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I just started reading the book by Edwards. I am currently in a complex variables course so i figured that I would give zeta a shot. I realize that there are easier ways to evaluate it for integer values, namely, the infinite sum. But I trying to at least evaluate it so that I could at least begin understanding the book. So, I was wondering if anyone could help me evaluate Riemann's integral form of zeta that is valid for all complex numbers
[tex]\zeta (2) = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \oint\limits_\gamma {\frac{x}{{{e^x} - 1}}} dx = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \left( {\int\limits_{ + \infty }^0 { + \int\limits_{x = \left| \delta \right|} + \int\limits_0^{ + \infty } {\frac{x}{{{e^x} - 1}}} } dx} \right)[/tex]
For instance, for s=2 I was trying to compute the integrals separately until I realized that the gamma integral was divergent for negative values of s. So, I figured that the contour integral part must turn it into some kind of indeterminate form, and I used the identity
[tex]\Gamma (1 - s) = \frac{\pi }{{\sin (\pi s)\Gamma (s)}}[/tex]
then,
[tex]\zeta (s) = \frac{{{e^{\pi is}}}}{{2i\sin (\pi s)\Gamma (s)}}\oint\limits_\gamma {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}} dx[/tex]
I got to here and plugged in gamma(2)=1, then realized that it really didn't make sense for integer values because sin(pi*n)=0 (which kind of made sense because the the previous form was divergent). So, to try to get this into an indeterminate or something I started working on the contour
[tex]\zeta (2) = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \oint\limits_\gamma {\frac{x}{{{e^x} - 1}}} dx = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \left( {\int\limits_{ + \infty }^0 { + \int\limits_{x = \left| \delta \right|} + \int\limits_0^{ + \infty } {\frac{x}{{{e^x}}}} } dx} \right)[/tex]
Sorry about the abuse of notation. I obviously didn't get very far because 1) I don't understand how this is a contour integral since x is a real variable and 2) I have no clue how to integrate x/(e^x-1).
[tex]\zeta (2) = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \oint\limits_\gamma {\frac{x}{{{e^x} - 1}}} dx = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \left( {\int\limits_{ + \infty }^0 { + \int\limits_{x = \left| \delta \right|} + \int\limits_0^{ + \infty } {\frac{x}{{{e^x} - 1}}} } dx} \right)[/tex]
For instance, for s=2 I was trying to compute the integrals separately until I realized that the gamma integral was divergent for negative values of s. So, I figured that the contour integral part must turn it into some kind of indeterminate form, and I used the identity
[tex]\Gamma (1 - s) = \frac{\pi }{{\sin (\pi s)\Gamma (s)}}[/tex]
then,
[tex]\zeta (s) = \frac{{{e^{\pi is}}}}{{2i\sin (\pi s)\Gamma (s)}}\oint\limits_\gamma {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}} dx[/tex]
I got to here and plugged in gamma(2)=1, then realized that it really didn't make sense for integer values because sin(pi*n)=0 (which kind of made sense because the the previous form was divergent). So, to try to get this into an indeterminate or something I started working on the contour
[tex]\zeta (2) = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \oint\limits_\gamma {\frac{x}{{{e^x} - 1}}} dx = {\left. {\frac{1}{{2i\sin (\pi s)}}} \right|_{s = 2}} \cdot \left( {\int\limits_{ + \infty }^0 { + \int\limits_{x = \left| \delta \right|} + \int\limits_0^{ + \infty } {\frac{x}{{{e^x}}}} } dx} \right)[/tex]
Sorry about the abuse of notation. I obviously didn't get very far because 1) I don't understand how this is a contour integral since x is a real variable and 2) I have no clue how to integrate x/(e^x-1).
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