1.4. HADAMARD’S “FINITE PART” METHOD 7 Exercise 1.17. Show that for any real numbers λ = 0 and μ and any holomor- phic function f(z) = λ z + μ z2 + o(z2) in a neighborhood of zero, with the notation of (1.13) and for any Schwartz function φ on R+ we have fpz=0 ˜ a+z (φ) = ∞,Riesz 0 xa φ(x) dx − δa+k μ φ(k−1)(0) (k − 1)! . Hint: Notice that 1 f(z) = 1 λ z − μ z + o(z). Applying the previous proposition to φ(x) = e−x leads to the following Riesz extension of the Gamma function. Given a complex number b, the map z → ˜ b + z) is meromorphic on the plane with simple poles in −b − N ∪ {0}. The finite part of the integral ˜ b + z), which is defined by the constant term in the Laurent expansion (1.14) ΓRiesz(b) := lim z→0 ˜ b + z) − 1 z Resz=0˜ b + z) , coincides with the ordinary Gamma function Γ(b) whenever Re(b) 0. Exercise 1.18. Show the following: (1) If b / −N ∪ {0}, then ΓRiesz(b) = ˜ b) = Γ(b + k) b (b + 1) · · · (b + k − 1) for any k such that Re(b) + k 0. (2) For any nonpositive integer b = −k show that ΓRiesz(0) = −γ ΓRiesz(−k) = Resb=−k˜ b) k j=1 1 j − (−1)k k! γ if k 0. Hint: Apply Proposition 1.16 to a = −(k + 1) and φ(t) = e−t. 1.4. Hadamard’s “finite part” method As before, φ denotes a Schwartz function on R+. Clearly, for Re(a) −1, Fa(φ) := ∞ 0 xa φ(x) dx = lim →0 ∞ xa φ(x) dx. In particular, for Re(b) 0, we have Γ(b) = lim →0 ∞ xb−1 e−x dx. Following Hadamard (see e.g. [Sch]) we want to extend Fa(φ) to all complex values of a in picking out the constant term denoted by ∞,Had 0 xa φ(x) dx in the asymp- totic expansion of the map → ∞ xa φ(x) dx as → 0, a procedure we are about to describe. Applying this to φ(x) = e−x yields an alternative extension ΓHad(b) := ∞,Had 0 xb−1 e−x dx of the Gamma function to the whole complex plane.

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