Help evaluating the zeta function

[galoisjr]

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

\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)

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

\Gamma (1 - s) = \frac{\pi }{{\sin (\pi s)\Gamma (s)}}

then,

\zeta (s) = \frac{{{e^{\pi is}}}}{{2i\sin (\pi s)\Gamma (s)}}\oint\limits_\gamma {\frac{{{x^{s - 1}}}}{{{e^x} - 1}}} dx

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

\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)

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).

 

[galoisjr]

So I finally figured out how to evaluate the integral which takes care of the hard parts I guess I still do not know what to do about the integral around the circle of radius delta though. I'm posting it for anyone who may need some insight because the integral is nasty and one little negative sign that you have to put in is what makes it so difficult. So far,

\begin{array}{l}<br /> \int {\frac{x}{{{e^x} - 1}}} dx = - \int {\frac{x}{{1 - {e^x}}}} \\<br /> u = x\\<br /> du = dx\\<br /> dv = \frac{1}{{1 - {e^x}}}<br /> \end{array}

to find the antiderivative of this for integration by parts we need to do more substitutions so,
\begin{array}{l}<br /> \int {\frac{1}{{1 - {e^x}}}} \\<br /> t = - {e^x}\\<br /> dt = - {e^x}dx\\<br /> dx = \frac{{dt}}{t}\\<br /> \int {\frac{{dx}}{{1 - {e^x}}}} = \int {\frac{{dt}}{{t(1 - t)}}} = \int {\frac{{dt}}{t} + \int {\frac{{dt}}{{1 - t}}} } \\<br /> w = 1 - t\\<br /> dw = - dt\\<br /> = \int {\frac{{dt}}{t}} - \int {\frac{{dw}}{w} = \log (t) - \log (w) = x - \log (1 - {e^x})} \\<br /> v = x - \log (1 - {e^x})<br /> \end{array}

Now we can solve the original integral using integration by parts and we have

\int {\frac{x}{{{e^x} - 1}}} dx = - \int {\frac{x}{{1 - {e^x}}}} = - ({x^2} - x\log (1 - {e^x}) - \int {x - \log (1 - {e^x})dx = x\log (1 - {e^x}) - \frac{{{x^2}}}{2} - \int {\log (1 - {e^x})dx} }

Now I didn't do this last integral because it is -Li_2(e^x) which is an infinite sum, so I felt like this was enough of testing my integration techniques

\left. {x\log (1 - {e^x}) - \frac{{{x^2}}}{2} + L{i_2}({e^x})} \right|_0^{ + \infty } = \frac{{{\pi ^2}}}{6}

Like I said before, I still do not know what to do with the integral around the half circle. So I still need some help If anyone would like to share some knowledge

 

[galoisjr]

It seems that I over looked the fact that gamma(1-s) has poles at the positive integers. So I guess that the second integral from 0 to infinity is sufficient for the evaluation of zeta at positive integers