This book deals with several aspects of what is now called "explicit number theory." The central theme is the solution of Diophantine equations, i.e., equations or systems of polynomial equations which must be solved in integers, rational numbers or more generally in algebraic numbers. This theme, in particular, is the central motivation for the modern theory of arithmetic algebraic geometry. In this text, this is considered through three of its most basic aspects. The local aspect, global aspect, and the third aspect is the theory of zeta and L-functions. This last aspect can be considered as a unifying theme for the whole subject.

integral we make the change of variables t = eu , and we obtain N r1 = N 1 B1 ({t}) sin(x log t) dt = tα log N sin(xu)e(1−α)u du . 0 It is now an easy exercise (for instance by explicit computation) to show that for 0 < α 1 the integral does not converge as N → ∞, and so neither does our series (there are, of course, other ways to prove this, for instance by grouping terms such that exp(kπ/|x|) m < exp((k + 1)π/|x|)). 9.3 Applications to Numerical Integration It is clear that in the

(2). We prove by induction on k that vp (Bk (χ)) z for all k 0. Let k 0, and assume that we have shown z for j < k, so that vp (Bk−j (χ)) z for 1 j k. Since that vp (Bj ) 1 and p | N , we have vp (N j /(j + 1)) 0, so trivially vp (j + 1) j for j the valuation of the second sum in the induction formula is also greater than or equal to z. For the ﬁrst term we consider three cases. Case 1: p f and f > 1. Then N = pf , and writing r = qp + s we have χ(r)rk = 0 r

Euler–MacLaurin summation formula; see Section 9.2.5. A more sophisticated, but easier to generalize, way of ﬁnding the value of the constant A has been explained in Section 9.2.5: by derivation of the formulas of Proposition 9.2.13, which come from the Euler–MacLaurin summation formula, we ﬁnd that log(n!) = (n + 1/2) log(n) − n − ζ (0) + O(1/n) . The value of ζ (0) is immediate to compute from the functional equation for the zeta function, which itself is a simple application of the Poisson

225 2 45 4 3 I(5, 3) = π ζ(7) − π ζ(5) − π 6 ζ(3) . 8 16 4 I(3, 5) = − 106. For k ∈ Z 1 ∞ Sk = , , set (log(2 sinh(t)) − t)k dt ∞ and Ck = 0 (log(2 cosh(t)) − t)k dt . 0 (a) By a series of successive changes of variable, or using Proposition 9.6.43 in a manner similar to the previous exercise, show that Sk = (−1)k k!ζ(k + 1)/2. (b) Show that C1 = ζ(2)/4 = π 2 /24 and C2 = ζ(3)/8 (note that for k 3 the expression for Ck involves Lik+1 (1/2), which is believed not to have any

Schl¨ omilch: tn Jn (x) = e(x/2)(t−1/t) . n∈Z (c) By multiplying this series with the one in which t is changed into 1/t, prove the following identities, valid for all x ∈ C. For all N ∈ Z=0 : Jn (x)Jn+N (x) = 0 , n∈Z and J02 (x) + 2 Jn2 (x) = 1 . n 1 This √ shows in particular that for all x ∈ R we have |J0 (x)| 1/ 2 for n ∈ Z=0 . 1 and |Jn (x)| 112. (a) Similarly to the ﬁrst question of the preceding exercise, show that for ﬁxed x, as ν → +∞ we have Yν (x) ∼ − (x/2)−ν Γ(ν) ∼− π 2 πν