1.3. RIESZ REGULARISATION 5 The following proposition gives the asymptotics at negative integers. Proposition 1.14. At negative integers −k −N, we have (1.12) ˜ −k + z) ∼0 (−1)k k! 1 z γ + k j=1 1 j . Proof. By formula (1.9) we have ˜ −k + z) = ˜ z) (z k) · · · (z 1) ∼0 1 z γ (z k) · · · (z 1) ∼0 (−1)k k! 1 z γ 1 + z k · · · (1 + z), from which (1.12) follows. Remark 1.15. Formula (1.12) yields back the residue at z = −k given by formula (1.10). 1.3. Riesz regularisation We now want to make sense of ˜ −k (φ) in spite of the fact that a = −k arises as a pole in (1.7). Following Riesz (see e.g. [Sch]), we pick the constant term at z = 0 in the Laurent expansion of the map z ˜ −k+z (φ). Proposition 1.16. Given a Schwartz function φ on R+ and a complex number a, the map z Fa+z(φ) is meromorphic on the plane with simple poles in −a N. The constant term in the Laurent expansion3 ∞,Riesz 0 xa φ(x) dx := fpz=0 ˜ a+z (φ) := lim z→0 ˜ a+z (φ) 1 z Resz=0 ˜ a+z (φ) (1.13) coincides with the ordinary integral 0 xa φ(x) dx whenever Re(a) −1. (1) If a / −N, then ∞,Riesz 0 xa φ(x) dx = ˜ a (φ) = (−1)k (a + 1) · · · (a + k) Fa+k(φ(k)), where k is any integer such that Re(a) + k −1. (2) If a = −k −N, then ∞,Riesz 0 x−k φ(x) dx = 1 (k 1)! 0 log x φ(k)(x)dx + Resa=−k ˜ a (φ) k−1 j=1 1 j , setting the sum over j equal to zero if k = 1. Proof. The case a / −N follows from the previous discussion. We therefore prove the proposition when a = −k for some integer k N. 3 Here fp stands for the finite part.
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