1.4. HADAMARD’S “FINITE PART” METHOD 11 Iterating the integration by parts procedure gives x−k φ(x) dx = 1 (k 1)! x−1 φ(k−1)(x) dx + −k+1 k 1 φ( ) + · · · + −k+j (k 1) · · · (k j) φ(j−1)( ) + · · · + −1 (k 1)! φ(k−2)( ) = 1 (k 1)! log x φ(k)(x) dx log (k 1)! φ(k−1)( ) + −k+1 k 1 φ( ) + · · · + −k+j (k 1) · · · (k j) φ(j−1)( ) + · · · + −1 (k 1)! φ(k−2)( ). (1.16) A Taylor expansion around = 0 of φ and its derivatives then gives rise to the finite part ∞,Had 0 x−k φ(x) dx = 1 (k 1)! 0 log x φ(k)(x) dx + φ(k−1)(0) (k 1)!(k 1) + · · · + φ(k−1)(0) (k 1) · · · (k j) (k j)! + · · · + φ(k−1)(0) (k 1)! = 1 (k 1)! 0 log x φ(k)(x) dx + φ(k−1)(0) (k 1)! k−1 j=1 1 k j . Exercise 1.22. Show that for any positive real number λ = we have fp =0 λ xa φ(x) dx = ∞,Had 0 xa φ(x) dx δa+k μ φ(k−1)(0) (k 1)! . Compare with Exercise 1.17. Applying Proposition 1.21 to φ(x) = e−x leads to the following Hadamard extension of the Gamma function (which we also call cut-off Gamma function): ΓHad(b) := fp =0 xb−1 e−x dx . Exercise 1.23. (1) Show that if Re(b) 0, the map xb−1 e−x dx lies in A[ ]. The corresponding asymptotic expansion has constant term given by ΓHad(b) = lim →0 xb−1 e−x dx = Γ(b). (2) Show that for b / Z≤0, the map xb−1 e−x dx lies in C A[ ] b . The corresponding asymptotic expansion has constant term given by ΓHad(b) = Γ(b + k) b(b + 1) · · · (b + k 1) .
Previous Page Next Page