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December 10

Reimann's integral from infinity to infinity

The question I am about to ask has been asked on wikipedia before but the answer seemed quite undecided and there was a lot of debate.

In "On the Number of Prime Numbers less than a Given Quantity.", Reimann makes a large jump between this step;

$\Pi (s-1)\zeta (s)=\int \limits _{0}^{\infty }{\frac {x^{s-1}dx}{e^{x}-1}}$

and this one;

$2\sin(\pi s)\Pi (s-1)\zeta (s)=i\int \limits _{\infty }^{\infty }{\frac {(-x)^{s-1}dx}{e^{x}-1}}$

Is this using a contour integral? (Perhaps a hankel contour would make sense in context? This paper seems to think so; http://www.damtp.cam.ac.uk/user/md327/fcm_3.pdf but seems to arrive at a slightly different result) or is it something else as it appears User:Eric Kvaalen was suggesting?

The original conversation can be seen here;

http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2012_August_22

Anyone know? — Preceding unsigned comment added by 5.81.8.11 (talk) 20:22, 10 December 2013 (UTC)[reply]

Presumably you integrate from infinity to infinity around a contour containing the pole at the origin. Sławomir Biały (talk) 15:44, 11 December 2013 (UTC)[reply]

I would have thought so which is why the Hankel contour appears to be the one to use but I cannot get it to arrive at that result! Help! — Preceding unsigned comment added by 5.81.14.188 (talk) 00:12, 12 December 2013 (UTC)[reply]