# An integral - Riemann's derivation of number of primes

• Nov 12th 2010, 05:43 PM
Kiwi_Dave
An integral - Riemann's derivation of number of primes
OK so I am not sure if this problem is analysis, number theory or calculus.

I am reading an English translation of Riemann's 1859 paper on the number of primes and cannot follow this bit.

He says consider this integral:

$\displaystyle \int^{+ \inf}_{+ \inf}\frac{(-x)^{s-1}dx}{e^x-1}$

From infinity in a positive sense around a domain that includes 0 but does not include any other point of discontinuity. This is then equal to:

$\displaystyle (e^{-\pi s \i}-e^{\pi s \i)\int^{+ \inf }_0 \frac{x^{s-1}dx}{e^x-1}$

I would like to fill in the details of this step but have been unsuccessful. I have tried:

Finding the residue at z=0 by solving:
$\displaystyle Res=\lim_{x \rightarrow 0} \frac{x(-x)^{s-1}}{e^x-1}=(-1)^{s-1} lim_{x \rightarrow 0}\frac{s\Pi(s-1)}{e^x}=(-1)^{s-1}s\int^{\inf}_0 \frac{x^{(s-1)}}{e^x}$

I also tried generalising x to z and integrating around the path from inf,$\displaystyle \epsilon$ to $\displaystyle -\epsilon,\epsilon$ to $\displaystyle -\epsilon,-\epsilon$ to $\displaystyle \inf,-\epsilon$ to $\displaystyle \inf,\epsilon$

where
$\displaystyle 0 < \epsilon < 2\pi$

This gave interesting results when $\displaystyle \epsilon$ either; tended to 0 equals pi or equals pi/2.

But non-the less I have not been able to get the required result. Is one of my approaches right? Is there a better approach?
• Nov 13th 2010, 11:54 AM
Drexel28
Quote:

Originally Posted by Kiwi_Dave
OK so I am not sure if this problem is analysis, number theory or calculus.

I am reading an English translation of Riemann's 1859 paper on the number of primes and cannot follow this bit.

He says consider this integral:

$\displaystyle \int^{+ \inf}_{+ \inf}\frac{(-x)^{s-1}dx}{e^x-1}$

From infinity in a positive sense around a domain that includes 0 but does not include any other point of discontinuity. This is then equal to:

$\displaystyle (e^{-\pi s \i}-e^{\pi s \i)\int^{+ \inf }_0 \frac{x^{s-1}dx}{e^x-1}$

I would like to fill in the details of this step but have been unsuccessful. I have tried:

Finding the residue at z=0 by solving:
$\displaystyle Res=\lim_{x \rightarrow 0} \frac{x(-x)^{s-1}}{e^x-1}=(-1)^{s-1} lim_{x \rightarrow 0}\frac{s\Pi(s-1)}{e^x}=(-1)^{s-1}s\int^{\inf}_0 \frac{x^{(s-1)}}{e^x}$

I also tried generalising x to z and integrating around the path from inf,$\displaystyle \epsilon$ to $\displaystyle -\epsilon,\epsilon$ to $\displaystyle -\epsilon,-\epsilon$ to $\displaystyle \inf,-\epsilon$ to $\displaystyle \inf,\epsilon$

where
$\displaystyle 0 < \epsilon < 2\pi$

This gave interesting results when $\displaystyle \epsilon$ either; tended to 0 equals pi or equals pi/2.

But non-the less I have not been able to get the required result. Is one of my approaches right? Is there a better approach?

I'm not sure if I'll be able to help, but could you explain the contour in a little more detail?
• Nov 13th 2010, 12:25 PM
Opalg
It's hard to see what is going on here without knowing more of the context, but the expression $\displaystyle \int^{+ \inf}_{+ \inf}\frac{(-x)^{s-1}dx}{e^x-1}$ suggests to me that it means the integral of $\displaystyle \frac{(-z)^{s-1}}{e^z-1}$ around a contour that goes from $\displaystyle +\infty$ to $\displaystyle \varepsilon$ along the real axis, then goes clockwise all the way round a circle of radius $\displaystyle \varepsilon$, and finally back along the real axis from $\displaystyle \varepsilon$ to $\displaystyle +\infty$.

Notice that the two integrals going in opposite directions along the real axis will not cancel each other out, because $\displaystyle \frac{(-z)^{s-1}}{e^z-1} = \frac{e^{(s-1)\log(-z)}}{e^z-1}$, and $\displaystyle \log(-z)$ will move onto another branch when z goes round the origin.
• Nov 13th 2010, 01:01 PM
Kiwi_Dave
The contour is allowed to be any closed contour starting at x=inf that does not enclose any singularity except for the one at x=0. I only described a contour using $\displaystyle \epsilon$ to demonstrate that I have had a fair crack at solving this myself!

The exact wording from the paper is:

If one now considers the integral

$\displaystyle \int\frac{(-x)^{s-1}dx}{e^x-1}$

from +inf to +inf taken in a positive sense around a domain which includes the value 0 but no other point of discontinuity of the integrand in its interior, then ths is easily seen to be equal to

$\displaystyle (e^{-\pi s \i}-e^{\pi s \i})\int^{+ \inf }_0 \frac{x^{s-1}dx}{e^x-1}$

provided that, in the many-valued function $\displaystyle (-x)^{s-1}=e^{(s-1)log(-x)}$, the logarithm is determined so as to be real when x is negative.
• Nov 14th 2010, 11:57 PM
Kiwi_Dave
I have this sorted now. If we place our branch cut on the positive x axis then as we approach the positive x axis from the positive y side then:

$\displaystyle (-x)^{(s-1)}=e^{(s-1)(ln(x)+i\pi)}=x^{(s-1)}e^{i\pi(s-1)}=x^{(s-1)}e^{i\pi(s-1)}(1)^{(s-1)}$

$\displaystyle =x^{(s-1)}e^{i\pi(s-1)}(e^{-2\pi i})^{(s-1)}=x^{(s-1)}e^{-i\pi(s-1)}$

Similarly as we approach the positive x axis from the negative y side then

$\displaystyle (-x)^{(s-1)}=e^{(s-1)(ln(x)+3i\pi)}=x^{(s-1)}e^{3i\pi(s-1)}=x^{(s-1)}e^{3i\pi(s-1)}(1)^{(s-1)}$

$\displaystyle =x^{(s-1)}e^{i\pi(s-1)}(e^{-2\pi i})^{(s-1)}=x^{(s-1)}e^{i\pi(s-1)}$

Hence the integral becomes:

$\displaystyle \int ^{+inf} _0 \frac {x^{(s-1)}e^{-i\pi(s-1)}dx}{e^x-1}+\int ^0 _{+inf} \frac {x^{(s-1)}e^{i\pi(s-1)}dx}{e^x-1}=(e^{-i\pi(s-1)}-e^{i\pi(s-1)}) \int ^{inf} _0 \frac {x^{(s-1)}dx}{e^x-1}$
• Nov 15th 2010, 01:10 AM
xxp9
How do you define $\displaystyle e^{i \pi (s-1)}$ and $\displaystyle e^{-i \pi (s-1)}$ ? From your first paragraph, these two are equal since
$\displaystyle x^{(s-1)}e^{i \pi (s-1)} = x^{(s-1)}e^{-i \pi (s-1)}$, but similarly to your computation we can conclude
$\displaystyle e^{i \pi (s-1)} e^{-i \pi (s-1)} = 1$. Then they must equal to 1 or -1, obviously ridiculous.
• Nov 15th 2010, 04:41 AM
shawsend
Quote:

Originally Posted by Kiwi_Dave
Is there a better approach?

Absolutely. Replace the hankel contour with it's mirror image and by doing so, elimiate the awkward $\displaystyle (-x)^{s-1}$ in the numerator and simultaneously convert the integration to the more "normal" principal branch-cut integral along the negative real axis. When you do this, you obtain the integral:

$\displaystyle \displaystyle\mathop\int\limits_{\multimap}\frac{z ^{s-1}}{e^{-z}-1}dz$

Now make the usual substitutions for a branch-cut integral over the principle branch-cut of log(z) along the negative real axis: starting along the bottom trace at $\displaystyle -\infty$, let $\displaystyle z=re^{-\pi i}$, around the origin, let $\displaystyle z=\rho e^{it}$ and along the top trace, let $\displaystyle z=re^{\pi i}$. Also, don't forget, you need to analyze this under the assumption that $\displaystyle \text{Re}(s)>1$, then show that as $\displaystyle \rho\to 0$ the integral around the origin goes to zero.