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Can you please demonstrate how one would calculate the Riemann Zeta function of any complex number, given that the Riemann Zeta function is equal to the following (shown in http://arxiv.org/pdf/1208.3429v1.pdf): enter image description here

If you utilize a technique in complex analysis (Such as Cauchy's Integral Formula), may you please explain the process step-by-step.

Thank you,

Best Regards,

J.M

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  • $\begingroup$ The technique used in this answer can be used for negative integers pretty easily math.stackexchange.com/questions/728590/… $\endgroup$ – ClassicStyle Jul 29 '14 at 5:50
  • $\begingroup$ So, you would use the Residue Theorem? (I thought it deals with branch cuts?) @TylerHG $\endgroup$ – VectorCalculus Jul 29 '14 at 5:57
  • $\begingroup$ Yes, the residue theorem is a nice approach for this...However, the cauchy integral formula could be used after reducing the integral to a series as seen in that example. Branch cuts are an unrelated topic though they are sometimes necessary when computing contour integrals. mathworld.wolfram.com/BranchCut.html $\endgroup$ – ClassicStyle Jul 29 '14 at 6:28
  • $\begingroup$ Thank you so much!! Can you please demonstrate what you mean (I understand the concept, however how does one apply Cauchy's integral theorem to a series?). @TylerHG $\endgroup$ – VectorCalculus Jul 29 '14 at 17:59
  • $\begingroup$ Sure, I'll post a solution later tonight. Thing to notice though is that the contour integral of a sum is the sum of contour integrals for each term. $\endgroup$ – ClassicStyle Jul 29 '14 at 18:44
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Sorry for the late response, I have been busy with finals and such.

Just for an example, let's calculate $\zeta(-1)$.

$$ \zeta(-1)=\frac{\Gamma(2)}{2\pi i}\oint_{\gamma}\frac{u^{-2}}{e^{-u}-1}du $$

Where $\gamma$ is the Hankel contour. Let's just focus on the integrand right now and expand it into a series. The full derivation on how this can be expanded is found in the link I provided in the comments. Thus:

$$ \frac{u^{-2}}{e^{-u}-1}=-\frac{1}{u^3}-\frac{1}{2u^2}-\frac{1}{12u}+\dots $$

Now, we can integrate this series term by term. Essentially, we are coming in from $-\infty$and going around the unit circle and then back out again to $-\infty$. The two paths coming in and out will cancel each other, as the only pole is at $u=0$. Thus, we can just focus on the unit circle contour. In general, the contour integral around the unit circle of the function $1/z^n$ is zero unless $n=1$. Thus,

$$\oint_{\gamma} \frac{u^{-2}}{e^{-u}-1}du=-\frac{1}{12}\oint_{\gamma}\frac{du}{u} $$

Then, by the Cauchy integral formula or the Residue theorem,

$$\oint_{\gamma}\frac{du}{u}=2\pi i $$

So,

$$ \zeta{(-1)}=\frac{-\Gamma(2)}{12(2\pi i)}2\pi i=-\frac1{12} $$

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  • $\begingroup$ Thank you Tyler, this is very concise!! Is this path counterclockwise or clockwise (I assume it is clockwise, as you applied the Cauchy Integral formula.) Also, what did you use to define the radius of the circle (I presume you can use anything and conventionally r, however I just want to be definite.) Lastly, seeing that the two paths canceled, is the distance between the two line contours 0. In addition, do the other terms in the contour integral of the series cancel because they are raised to different powers and not 1? Thank you so much Tyler, I wish you well in your finals!! @TylerHG $\endgroup$ – VectorCalculus Jul 31 '14 at 3:53
  • $\begingroup$ The hankel contour is counterclockwise, and The cauchy integral formula is defined for counterclockwise traversals. But it is only a sign change depeding on which way you go. We are free to change the radius of the circle as long as a pole isn't crossed when doing this. Yes, the two paths are a zero "distance" apart. Yes, the other terms in the series cancel out because they are raised to powers other than one. It's no problem. If you want to read a book on complex analyis "Visual complex analysis" is a great introduction to the subject. You are welcome! It was my pleasure, cheers! $\endgroup$ – ClassicStyle Jul 31 '14 at 4:04
  • $\begingroup$ Hello Tyler, I am just confirming that u is a complex variable, seeing that you applied these methods. @TylerHG $\endgroup$ – VectorCalculus Jul 31 '14 at 20:15
  • $\begingroup$ Yes, $u$ is a complex variable...none of this would make sense if it wasnt $\endgroup$ – ClassicStyle Jul 31 '14 at 20:16
  • $\begingroup$ Sorry to bother you again Tyler, but I calculated zeta of -3 and it had an outcome of -11/30 or -66/180, when it should be -1/120. Do you know what I am doing wrong in the series expansion?@TylerHG $\endgroup$ – VectorCalculus Aug 1 '14 at 16:16

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