[SOLVED] Von-Mangoldt Explicit Formula

**raheel88** · April 25th 2010, 05:15 AM

Does anyone know of a book/website where I can find a proof of the Von-Mangoldt Explicit Formula? Namely;

$\psi (x) = x - \displaystyle\sum_{\rho}\frac{x^\rho}{\rho}\ - \frac{1}{2}\ ln(1 - \frac{1}{x^{2}}) - ln(2\pi)\$

where the rhos are the zeta zeros whose real parts are less than 1.

Thanks in advance!

**chisigma** · April 25th 2010, 06:02 AM

Originally Posted by raheel88

Does anyone know of a book/website where I can find a proof of the Von-Mangoldt Explicit Formula? Namely;

$\psi (x) = x - \displaystyle\sum_{\rho}\frac{x^\rho}{\rho}\ - \frac{1}{2}\ ln(1 - \frac{1}{x^{2}}) - ln(2\pi)\$

where the rhos are the zeta zeros whose real parts are less than 1.

Thanks in advance!

All zeroes of the $\zeta(s)$ have real part less than 1!... the term $\displaystyle\sum_{\rho}\frac{x^\rho}{\rho}\$ is the contribute to $\psi (x)$ of the complex zeroes of $\zeta (s)$ ...

Kind regards

$\chi$ $\sigma$

**raheel88** · April 25th 2010, 06:18 AM

I see! And taking a taylor expansion of the $- \frac{1}{2}\ ln(1 - \frac{1}{x^{2}})$ term yields the contribution from the trivial zeta zeroes?

But what significance do the other two terms have? Namely the $x$ and the $-ln(2\pi)$ ?

Many thanks!

**shawsend** · April 25th 2010, 07:12 AM

I do not know of a book or web site that does a clear job of explaining the proof. Start with Perron's Formula:

$\psi(x)=-\frac{1}{2\pi i}\mathop\int\limits_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\fra c{x^s}{s}ds$ .

and for the record that's true if x is not a prime power. If it is, need a 1/2 log(p) on the left side.

As you can see, the singular points are at the zeta zeros, the zeta pole at one, and zero right. Now take a square contour pinned at Re(s)=c and allowed to expand without bounds and always going between the zeros of the zeta function although I think an argument could be made for going right through the zeros but I digress. It's tedious proving the integral over the two horizontal legs and the left vertical leg goes to zero as the contour grows without bounds. So assuming it does, then we just sum up the residues: The infinite sum comes from the non-trivial zeros, the log(1-1/x^2) comes from the trivial ones, x comes from the residue at one, and the log(2pi) comes from the residue at zero.

**chiph588@** · April 25th 2010, 08:33 AM

Originally Posted by shawsend

I do not know of a book or web site that does a clear job of explaining the proof. Start with Perron's Formula:

$\psi(x)=-\frac{1}{2\pi i}\mathop\int\limits_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\fra c{x^s}{s}ds$ .

This might not be to hard to prove actually.

Given $\Re(s)>1$ , then $-\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\Lambda(n)}{n^s} = \int_{0}^{\infty} \psi(x)\frac{s}{x^{s+1}}dx$ where the last equality comes from partial summation.

If we perform the substitution $x=e^t$ , we can then apply the inverse Laplace transform to get what we want. I suppose the problem with this is knowing why the inverse Laplace transform formula is true.

**chiph588@** · April 27th 2010, 08:45 PM

Originally Posted by shawsend

I do not know of a book or web site that does a clear job of explaining the proof. Start with Perron's Formula:

$\psi(x)=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\fra c{x^s}{s}ds$ .

First off, I claim that given $c>1 \text{ and } x\in\mathbb{R}$ , $\displaystyle \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{e^{xs}}{s}ds = \begin{cases} 0, & \mbox{if } x<0 \\ \tfrac12, & \mbox{if } x=0 \\ 1, & \mbox{if } x>0 \end{cases}$ . . (1)

This is not too hard to prove assuming you choose the right contour to integrate over. I'm omitting the proof but ask if you'd like to see it.

Define $\displaystyle f(s) = -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^\infty \Lambda(n)n^{-s}$ , $\Re(s)>1$ .

Now $\displaystyle \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{f(s)e^{xs}}{s}ds = \sum_{n=1}^\infty \frac{\Lambda(n)}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{(x-\log(n))s}}{s}ds$ .

But by (1) , $\displaystyle \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{(x-\log(n))s}}{s}ds = \begin{cases} 0, & \mbox{if } x-\log(n)<0 \\ \tfrac12, & \mbox{if } x-\log(n)=0 \\ 1, & \mbox{if } x-\log(n)>0 \end{cases}$

Thus $\displaystyle \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \frac{f(s)e^{xs}}{s}ds = \sum_{\log(n) < x} \Lambda(n) + \frac12\Lambda(x)\beta(x)$ , where $\displaystyle \beta(x) = \begin{cases} 1, & \mbox{if }x\mbox{ is a prime power} \\ 0, & \mbox{else} \end{cases}$

Let $\displaystyle x\mapsto\log(x)$ to see $\displaystyle -\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\fra c{x^s}{s}ds = \sum_{n \leq x} \Lambda(n) - \frac12\Lambda(x)\beta(x)= \psi_0(x)$

**chiph588@** · April 27th 2010, 09:14 PM

As you can see from above, we hit a little snag when the integral has the value of $\frac12$ . This is why "The Chebyshev smoothing function" was introduced in the original proof of the prime number theorem.

$\psi_1(x) = \int_{0}^x \psi(u)du = \int_{1}^x \psi(u)du = \sum_{n \leq x} (x-n)\Lambda(n)$

It turns out that $\psi_1(x) = -\frac{1}{2\pi i} \int_{c-\infty i}^{c+\infty i} \frac{\zeta'(s)}{\zeta(s)}\frac{x^{s+1}}{s(s+1)}ds$ which is a bit nicer than what's above.

**raheel88** · April 28th 2010, 03:29 AM

hi chiph588,

thanks for the tips. analytic number theory is one tricky, but deeply interesting, branch of mathematics...i've only recently been introduced to it and i'm glad i did!

i spoke to my lecturer about this and he explained an almost identical approach (inverse mellin transform of the V.M. formula to get psi(x)) but your explanation is very thorough and helped me alot more.

thanks again.

regards

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