Riemann Prime Counting Function

1. Sep 3, 2006

saltydog

Can anyone tell me if Riemann's Prime Counting function can be solved by residue integration?

Here it is:

$$J(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{ln(\zeta(s))x^s}{s}ds$$

which has the solution:

$$J(x)=li(x)-\sum_{\rho}li(x^\rho)-ln(2)+ \int_x^{\infty}\frac{dt}{t(t^2-1)ln(t)}$$

I mean it has four sets of poles: 0,1, trivial zeros, complex zeros (due to log term) and I would suspect each of the four terms above may result from the residues there assuming the integral goes to zero around the remainder of a closed contour.

Anyway I'll be looking into it as well.

Last edited: Sep 3, 2006
2. Sep 4, 2006

lokofer

I think the formula for J(x) can be derived from the "Von Mangoldt formula for $$\psi (x)$$ since:

$$log\zeta(s) = \int_{2}^{\infty} \frac{x}{log(x)}x^{-s}d\psi$$

and the formula for $$\psi (x)$$ was already known to Riemann..

3. Sep 4, 2006

saltydog

Thanks Lokofer. I'm stuborn though.

Suppose I wanted to calculate the residue at the first complex zero of Zeta for the integrand above. That then would be:

$$\mathop\text{Res}\limits_{z=\rho_1}\left\{\frac{ln(\zeta(z))x^z}{z}\right\}= \frac{1}{2\pi i}\oint\frac{ln(\zeta(z))x^z}{z}dz$$

I'd resort to this because I don't see how to calculate it any other way. However, the log term in the expression causes a problem I think because of the branch-cut along the negative x-axis. I thought though that I wasn't integrating through the branch cut but if you plot zeta along the circular contour around the first zero, it cuts across the negative x-axis which I guess cuts across the branch-cut of the log function. I'm not very clear about this though.. The plot below is the Re and Im components of zeta around this contour. Note at around 3 it's cutting through the negative x-axis.

See, this is precisely why I need to take a course in all this. I just don't follow it too well.

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Last edited: Sep 4, 2006
4. Sep 4, 2006

shmoe

You can't avoid the branch cut. You can't define log zeta in a punctured neighbourhood of a zero and get something analytic.

5. Sep 4, 2006

saltydog

That makes sense. Thank you. I see the problem now, the whole problem indeed: need to crawl before I can walk. I'm impatient though. It's all very interesting to me.

6. Sep 4, 2006

shmoe

It will come. Diving at a hard problem and filling in gaps needed to understand it is not a bad way to go. At some point it does pay to go through a subject in detail to pick up anything you may have missed. I think you mentioned you're taking a complex analysis course this fall? That should be good, you have plenty of interesting problems already in your hand to apply stuff as you learn it.

7. Sep 4, 2006

saltydog

Yea, I'm taking a class however I don't feel comfortable asking my professor anything about this . . . he'll get the wrong impression and I'd just as soon remain unnoticed.

8. Sep 4, 2006

shmoe

It's every professors nightmare to have a student that's actually motivated, so you are right to keep quiet.

That was sarcasm by the way. If you are up to date with what he wants you to know in class, then there should be no problem with discussing more advanced stuff during office hours or after class (if he has time). You have some definite goals in the form of number theory applications you'd like to get out of his course, it's not a bad thing to tell him this. I know I would be thrilled to know a student had a motivation beyond "this course is required for my program".

Being unnnoticed isn't a good thing when it comes time to ask for letters of reccomendation.

9. Sep 4, 2006

lokofer

- the post by "saltydog" has made me create a doubt..what would happen with a more "general" complex integral such us:

$$\frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty}ds \frac{log \zeta (as)}{as}$$ ?

I think that you can this expression by setting $$x=u^{1/a}$$ inside the usual expression for J(x) however i'm not pretty sure.

According to this the last interal should read

$$\int_{u^{1/a}}^{\infty}dt \frac{1}{t(t^{2} -1)}$$

10. Sep 4, 2006

saltydog

In the interest of some closure in this matter I'll summarize Riemann's evaluation of the integral. He relies on the two expressions for the auxiliary function Xi:

$$\xi(s)=\pi^{-s/2}\Gamma(s/2+1)(s-1)\zeta(s)$$

and:

$$\xi(s)=\xi(0)\mathop\Pi\limits_{\rho}(1-\frac{s}{p})$$

Now, taking logarithms of both of these expressions, $ln(\zeta(s))$ can be extracted and expressed as a sum of 5 terms:

$$ln(\zeta(s))=ln(\xi(0))+\sum_{\rho}(1-\frac{1}{\rho})-ln(\Gamma(s/2+1))+\frac{s}{2}ln(\pi) -ln(s-1)$$

These terms can now be substituted into the integral expression for J(x) and the integration performed termwise (see Edwards).

Last edited: Sep 4, 2006
11. Mar 8, 2010

The_Shape

Here is something that may be of interest. The full title is:
"THE ANALYTIC EXPRESSION FOR RIEMANN'S PRIME COUNTING FUNCTION VIA THE RESIDUE THEOREM."

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