1.4. HADAMARD’S “FINITE PART” METHOD 9 Exercise 1.20. Deduce from the previous lemma that the map log x e−x dx lies in A[ ] A[ ] log , and that we have lim →0 log x e−x dx = −γ. Proposition 1.21. Let φ denote a Schwartz function on R+. (1) If Re(a) −1, the map xa φ(x) dx lies in C A[ ] a+1 , and we have lim →0 xa φ(x) dx = 0 xa φ(x) dx = Fa(φ). (2) If a / −N, the map xa φ(x) dx lies in C A[ ] a+1 with constant term given by ∞,Had 0 xa φ(x) dx = (−1)k (a + 1) · · · (a + k) 0 xa+k φ(k)(x) dx = ˜ a (φ) for any integer k such that a + k −1. (3) If a = −k −N, the map xa φ(x) dx lies in A[ ] −k+1 A[ ] log with constant term given by ∞,Had 0 x−k φ(x) dx = 1 (k 1)! 0 log x φ(k)(x) dx + Resa=−kFa(φ) k−1 j=1 1 j = ˜ −k (φ), where the sum is set to zero if k = 1. Proof. (1) If Re(a) −1, we have xa φ(x) dx = 0 xa φ(x) dx 0 xa φ(x) dx = 0 xa φ(x) dx a+1 1 0 xa φ(x) dx. The map 1 0 xa φ(x) dx lies in A[ ] so that the map xa φ(x) dx lies in C A[ ] a+1 and lim →0 xa φ(x) dx = 0 xa φ(x)dx. (2) If a / −N, we proceed by induction on k with Re(a) ] (k + 1), −k] using integration by parts to show that xa φ(x) dx lies in C⊕A[ ] a+1 . The step k = 0 holds by the previous item. One integration by parts provides the induction step. Indeed, if the map xa+1 φ(x) dx lies in C⊕A[ ] a+2 , then the map xaφ(x) dx lies in C⊕A[ ] a+1 since xa φ(x) dx = xa+1 a + 1 φ (x) dx + xa+1 a + 1 φ(x) = 1 a + 1 xa+1 φ (x) dx a+1 a + 1 φ( ).
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