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7, it is given by r β= l=1 2n n −2 l l p p , r= log 2n . 27) in mind, we see that each summand is 0 or 1, and so β ≤ 2n r ≤ log log p . 25), pr < (2n)π(2n)r log p < (2n)π(2n) log 2n. = p≤2n 2n n 2n , so that 2n n log 2n ≤ log ≤ π(2n) log 2n. 27), we obtain π(x) ≥ π 2 x 2 x 2 log 2 x . 28) to the effect that those prime p, n < p ≤ 2n do not divide the denominator, and therefore 2n p. Hence n must be divisible by n

11) ∞ −x2 e 0 dx = √ π 2 . 9), put t = sin2 θ to obtain π 2 B(α, β) = 2 sin2α−1 θ cos2β−1 θ dθ. 13) D x where D = x2 + y 2 ≤ X . By the change of variable x = y 0≤ r cos θ, y = r sin θ, we have the correspondence r θ ˜= D↔D 0 ≤ r ≤ X, 0 ≤ θ ≤ π 2 , where the absolute value of the Jacobian of this transformation is r. 12) above. 10). 11) follows. It follows that ∞ e− x2 2 dx = √ 2π, −∞ which is used to normalize the distribution function of the Gaussian (or normal) distribution. 13) 2 dx R = 4 lim re R→∞ 0 π 2 −r 2 2 1 = 4 − e−r 2 dθ dr 0 0 ∞ 0 π = π.

42) interpolates g(x) (L(αi ) = g(αi )), where n f (x) = i=1 (x − xi ). 4 (Carlitz [Car]) Let k1 , k2 , · · · , kn be pairwise relatively prime positive integers and let gi (x) = xk i − 1 = x−1 n−1 l=1 x − ζil , 1 ≤ i ≤ n, where ζi signifies a primitive ki -th root of 1. Further put n−1 fi (x) = (x − 1) gi (x) = xki − 1 = l=1 x − ζil and Gi (x) = gk (x), k=i 1 ≤ i ≤ n. 44) interpolates the polynomial φi (x) of degree < ki − 1 such that n Gi (x) φi (x) = 1. i=1 Proof. We note the following facts.

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Archimedean Zeta Integrals by Garett P.


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