I am reading the second chapter of Titchmarsh's book on the Riemann Zeta Function. I would have written:
ζ(12)=1+1√2+1√3+⋯=∞
If you think about it for a moment. This doesn't decay nearly fast enough, and so the sequence diverges. Then I had to look up the actual definition of ζ(s) in the region s=σ+it and 0<σ<1. We have:
ζ(s)={∑1nsRe(s)>1s∫∞0[x]−xxs+1dx1>Re(s)>0
Then if I evaluate at s=32=32+0i we use the second formula:
ζ(12)=12∫∞0[x]−xx3/2dx=∞∑n=0[2√n−2√n+1−1√n+1]?<0
Is this thing negative? Is ζ(12)<0. The book overs several "analytic continuations" and I'm only looking at this first one, to make sure I understand.
Could someone help me evaluate the integral? I didn't use any fancy changes of variables. The main step is:
∫∞0f(x)dx=∑∫n+1nf(x)dx=∫10f(x)dx+∫21f(x)dx+…
Answer
You can take advantage of the
following identity:
ζ(12)=N∑k=11√k−2√N−12∫∞Nx−⌊x⌋x3/2dx,N=1,2,3,…
Note that
0<|12∫∞Nx−⌊x⌋x3/2dx|<1√N such that you don't need to evaluate the integral. Namely,
\bbx{\zeta\pars{1 \over 2} = \lim_{N \to \infty}\pars{\sum_{k = 1}^{N}{1 \over \root{k}} - 2\root{N}}}
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