camilus
Jan25-11, 07:48 PM
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.
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.