>>14243653Riemann knew the reflection property of the (more or less) L function ?(s)?(s), and the fact that it's a Mellin transform of something nice. In fact, maybe you can try to figure out how the same Mellin transform, coupled with what you know about the poles of the Gamma function, gives
?(-n) = (-1)^n B_{n+1} /(n+1)
for n a positive integer. Since all odd Bernoulli numbers except B_1 are zero, this gives you lots of zeros to play around with.
Again, all of this you get from the Mellin transform of e^{-? n x}, and you sum over n. It's a nice geometric sum, and gives you all of the above.
What Riemann did was to look at the Mellin transform of e^{-? n^2 x}. You get more or less the same thing, but summing over n is less trivial. However, what's really important for Mellin transforms is how the thing behaves at 0 and infinity. Luckily, Riemann knows that the sums of x are related to sums of 1/x. He uses that to make the integral more tractable. He gets a really nice formula if he zooms in on the strip s = 1/2 + i t, because he gets an expression like something * (x^{ s-2 } + x^{-1-s} ), which simplifes the powers into just a cosine.
He says he can do the integral as a rapidly convergent series in t^2. This is what he did.
A legitimate question is: why e^{-? n^2 x}? It's because this lets you sum over all n in Z, and lets the Mellin integral converge. If you do e^{-? n x}, you can only sum over positive n, which is OK but it doesn't let you relate asymptotics at 0 to those at infinity, and you need this to simplify the integral.
These are some standard methods in asymptotic analysis.
