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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( ).