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

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

Thanks in advance!

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- Apr 25th 2010, 06:15 AMraheel88[SOLVED] Von-Mangoldt Explicit Formula
Does anyone know of a book/website where I can find a proof of the Von-Mangoldt Explicit Formula? Namely;

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

Thanks in advance! - Apr 25th 2010, 07:02 AMchisigma
- Apr 25th 2010, 07:18 AMraheel88
I see! And taking a taylor expansion of the term yields the contribution from the trivial zeta zeroes?

But what significance do the other two terms have? Namely the and the ?

Many thanks! - Apr 25th 2010, 08:12 AMshawsend
I do not know of a book or web site that does a clear job of explaining the proof. Start with Perron's Formula:

.

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. - Apr 25th 2010, 09:33 AMchiph588@
This might not be to hard to prove actually.

Given , then where the last equality comes from partial summation.

If we perform the substitution , 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. - Apr 27th 2010, 09:45 PMchiph588@
- Apr 27th 2010, 10:14 PMchiph588@
As you can see from above, we hit a little snag when the integral has the value of . This is why "The Chebyshev smoothing function" was introduced in the original proof of the prime number theorem.

It turns out that which is a bit nicer than what's above. - Apr 28th 2010, 04:29 AMraheel88
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