Riemann’s Zeta Function

I guess it is about time to get to the zeta function side of this story, if we’re ever going to use Farey sequences to show how you could prove the Riemann hypothesis. I’ve been reading a bit of Edwards’ book, with the same title as this post, and thought I’d try to summarize the first chapter over the next few posts. I’m not sure that the content of the first chapter is hugely vital for the final goal of relating to the Farey sequences, but I wanted to try to learn some of it anyway.

I should mention, before talking about any of this, that I will not claim to know any complex analysis. The last complex analysis course I took was 5 years ago, I can’t be sure how much I paid attention at the time, and I haven’t used it since. I will be jumping over gaps of various sizes for quite a while in the upcoming posts. Perhaps sometimes I’ll mention that a gap is there. Mostly, though, what I’m after in this story is the outline, and how all of the parts fit together, as a big picture. Perhaps I’ll go back and fill in some gaps, as I understand more.

For today, let’s see if I can build up to an analytic expression of \zeta(s). Our starting point is the function

\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}

which is defined for real values of s larger than 1. The goal is to find a nice expression that is defined on more of the complex plane, but agrees with this definition on the reals larger than 1.

To get to that point, we’ll use the \Gamma function:

\Gamma(s)=\displaystyle\int_0^{\infty} e^{-x}x^{s-1}\ dx

This is an analytic function defined everywhere except at the negative integers and 0. If you are only interested in real values of s, this function is a continuous extension of the factorial function, which is only defined on the positive integers. Actually, there is a shift involved, so \Gamma(n)=(n-1)! (Edwards blames this shift on Legendre, whose reasons, he states, “are obscure”. Edwards uses \Pi(s)=\Gamma(s+1) throughout, so I might make some \pm 1 errors). In particular, s\Gamma(s)=\Gamma(s+1) when both sides make sense.

Another relation that we’ll use is that

\pi=\Gamma(s)\Gamma(1-s)\sin(\pi s)

If memory serves (from reading Artin’s lovely little book on the \Gamma function), you can obtain this by determining that the derivative of the expression on the right is 0. This tells you that \sin(\pi s)\Gamma(s)\Gamma(1-s) is a constant, and so you can pick your favorite value of s to obtain the value of that constant. Anyway, the identity I’ll use is a rearrangement of the one above, namely

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

Now, in the expression

\Gamma(s)=\displaystyle\int_0^{\infty}e^{-x}x^{s-1}\ dx,

substitute x=nu for some positive n. Messing about with that substitution for a few minutes (and then re-writing u=x by abuse of notation), you can arrive at

\dfrac{\Gamma(s)}{n^s}=\displaystyle\int_0^{\infty} e^{-nx}x^{s-1}\ dx.

That n^s in the denominator is useful for us, as that’s how it shows up in \zeta(s). In particular, we can obtain

\begin{array}{rcl}\Gamma(s)\zeta(s) &=& \displaystyle \Gamma(s)\sum_{n=1}^{\infty}\dfrac{1}{n^s}=\sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s} \\ &=& \displaystyle \sum_{n=1}^{\infty}\int_0^{\infty}e^{-nx}x^{s-1}\ dx \\ &=& \displaystyle \int_0^{\infty}x^{s-1}\sum_{n=1}^{\infty} e^{-nx}\ dx\\ &=& \displaystyle \int_0^{\infty} \dfrac{x^{s-1}}{e^x-1}\ dx\end{array}

That last transition coming about by summing the geometric series. There are probably some things analysts like to check in here, moving infinite sums and improper integrals past each other… I’ll let them.

Ok, we’re nearly there. Next up, you do some complex line integral and show that

\begin{array}{rcl} \displaystyle \int_{+\infty}^{+\infty} \dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x} &=& \displaystyle (e^{i\pi s}-e^{-i\pi s})\int_0^{\infty}\dfrac{x^{s-1}}{e^x-1}\ dx \\ &=& 2i\sin(\pi s)\Gamma(s)\zeta(s)\end{array}.

That integral is a little weird, going “from +\infty to +\infty“. Really, we take a path that “starts at +\infty“, swings around 0, then goes back out to infinity. This is almost certainly one of the complex analysis things I should go back and learn more about. Since we are integrating (-x)^s=e^{s\log(- x)} along positive real values of x, we’re working with logs along the negative real axis. Perhaps I’ll return to this integral in a future post.

Using the identity about \sin(\pi s)\Gamma(s) from above, we obtain

\zeta(s)=\dfrac{\Gamma(1-s)}{2\pi i}\displaystyle \int_{+\infty}^{+\infty}\dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x}.

We’re now at a point where we’ve (apparently) got an expression for \zeta(s) that is analytic in the complex plane except for a simple pole at s=1.

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One Response to “Riemann’s Zeta Function”

  1. Another Formula for J(x) « ∑idiot's Blog Says:

    […] I introduced , I ended with the following […]

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