analytically continuing the Riemann zeta function (RZF) using the gamma function leads to this identify: \(n^{-s} \pi^{-s \over 2} \Gamma ({s \over 2}) = \int_0^{\infty} e^{-n^2 \pi x} x^{{s \over 2}-1} dx\) ________(1) and from that we can build a similar expression incorporating the RZF: \(\pi^{-s \over 2} \Gamma ({s \over 2}) \zeta (s) = \int_0^{\infty} \psi (x) x^{{s \over 2}-1} dx\) ________(2) where \(\psi (x) = \sum_{n=1}^{\infty} e^{-n^2 \pi x}\) is the Jacobi theta function. Then Riemann proceeds to use the functional equation for the theta function: \(2\psi (x) +1 = x^{-1 \over 2} (2 \psi ({1 \over x})+1)\) to equate (2) with: \(\pi^{-s \over 2} \Gamma ({s \over 2}) \zeta (s) = \int_1^{\infty} \psi (x) x^{{s \over 2}-1} dx + \int_1^{\infty} \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx+{1 \over 2} \int_0^{1} (x^{-1 \over 2} x^{{s \over 2}-1} - x^{{s \over 2}-1}) dx\) This is the step Im stuck on, Im trying to figure out what he did to get that last equation. Any help would be greatly appreciated.
It is after decomposing the integral over \((0,\infty)\) into the sum of integrals over \((0,1)\) and \((1,\infty)\), and using the functional equation. Actually the second integral in must go over \((0,1)\), not over \((1,\infty)\). The next step would be to change variables \(\frac1x\to x\) to transform the integrals over \((0,1)\) to ones over \((1,\infty)\).
there's no need to transform the third integral seeing as it evaluates to a very neat expression 1/[s(s-1)] explicitly showing the pole at s=1. as for the second integral, I kind of see what you're saying, I believe the approach is correct, but I havent got the expression to match: \(\int_1^{\infty} \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx = \int_1^{\infty} \psi (x) x^{-{s+1 \over 2}} dx\) using the functional equation.
What I was saying is that the second integral is not \(\int_1^{\infty} \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx.\) It is \(\int_0^1 \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx.\) The functional equation is already used. What you need now is change of variables under the integral
holy crap, I been on a wild goose chase because of a typo! Maybe I should let the publisher know.. I had to check Riemann's original handwritten manuscript to see that what you're saying is right and that the second integral is from 0 to 1.
problem resolved. I verified and indeed \(\pi^{-s \over 2} \Gamma ({s \over 2}) \zeta (s) = \int_1^{\infty} \psi (x) x^{{s \over 2}-1} dx + \int_0^1 \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx+{1 \over 2} \int_0^{1} (x^{-1 \over 2} x^{{s \over 2}-1} - x^{{s \over 2}-1} ) dx\) => \(\zeta (s) = {\pi^{s \over 2} \over \Gamma (\frac{s}{2})} \left( \int_1^{\infty} \psi (x) x^{{s \over 2}-1} dx + \int_0^1 \psi ({1 \over x}) x^{-1 \over 2} x^{{s \over 2}-1} dx+{1 \over 2} \int_0^{1} (x^{-1 \over 2} x^{{s \over 2}-1} - x^{{s \over 2}-1} ) dx \right)\) => \(\zeta (s) = {\pi^{s \over 2} \over \Gamma (\frac{s}{2})} \left( {1 \over s(s-1)} + \int_1^\infty \psi(x) \left( x^{{s \over 2}-1} + x^{-{s+1 \over 2}} \right) dx \right)\)
The error Im talking about can be found in the middle of page 3 not counting the cover page. http://www.claymath.org/millennium/Riemann_Hypothesis/1859_manuscript/EZeta.pdf Riemann's manuscript, bottom of page 2, where you can see the real version. http://www.claymath.org/millennium/Riemann_Hypothesis/1859_manuscript/riemann1859.pdf
Interesting. At least it is good that the bug is not in the official statement of the Millennium Prize problem Please Register or Log in to view the hidden image! For a moment, when I saw the URL of your link I was shocked it included a claymath address.
