By Graham Everest
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This completes the first proof for the analytic continuation of the zeta function into (s) > 0. ✷ Why was it necessary to split the integral from 1 to ∞ into all these subintegrals? Answer: The Taylor approximation in (37) is only valid for bounded ∞ values of h log(t). If we had stuck to 1 all the way, t would be arbitrary and the quantity h log(t) would be unbounded. By the splitting of the integral, we had only to consider t ∈ [n, n + 1] for a fixed n. These are treacherous waters! 3. GE’s method: Has the additional benefit of giving a continuation to the whole of the complex plane (with exception of the simple pole at s = 1).
But it really makes the functional equation for the zeta function work. Note that the series defining θ converges uniformly in the range y > δ for any fixed δ > 0 (see also exercise 30). 9 2 y2 fb (y) := f(by) = e−πb . 8 ˆfb (n). fb (n) = Z (53) Z What is ˆfb (y)? ˆfb (y) = fb (x)e−2πixy dx = R f(bx)e−2πixy dx. (54) R Now put u := bx, so dx = 1b du: Equation (54) becomes ˆfb (y) = 1 b y 1 y f(u)e−2πiu b du = ˆf . 9 to this equation, so ˆfb (y) = 1 f y . b b (56) Put this result into equation (53) and insert the definition of f again: 2 m2 e−πb Z Finally, put b := √ = 1 b m2 e−π b2 .
We have for all 0 < x < 1 x f(x) = 0 dt = 1 + t2 ∞ x (−t2 )n dt n=1 ∞ = n=0 0 (−1)n 2n+1 x , 2n + 1 (94) because there we may interchange integration and summation thanks to the uniform convergence. Now, we may take the limit x → 1 thanks to Abel’s Limit theorem (a nice special feature of power series - see Appendix B, since this is rather a theorem of Calculus). For x → 1, we get L(1, χ) on the right-hand side, and the integral in (92), f(1) = π/4 on the left-hand side. 19). Proof: Check all cases for m and n modulo 4 in χ(mn) = χ(m)χ(n) resp.