6 1. GAMMA FUNCTION EXTENDED TO NONPOSITIVE INTEGER POINTS (1) Let us start with k = 1. Integrating by parts for Re(z) −1, we have ˜ −1+z (φ) = − 1 z ∞ 0 xzφ (x)dx. Since Resz=0 ˜ −1+z (φ) = φ(0), we have lim z→0 ˜ −1+z (φ) − 1 z Resz=0 ˜ −1+z (φ) = lim z→0 − 1 z ∞ 0 xzφ (x)dx − φ(0) z = lim z→0 − ∞ 0 xz − 1 z φ (x) dx = − ∞ 0 log x φ (x)dx. (2) When k 1, integrating by parts k times, for Re(z) −1 we find that ˜ −k+z (φ) = (−1)k (−k + 1 + z) · · · (z − 1)z ∞ 0 xzφ(k)(x)dx. Since Resz=0 ˜ −k+z (φ) = φ(k−1)(0) (k−1)! , we have lim z→0 ˜ −k+z (φ) − 1 z Resz=0 ˜ −k+z (φ) = lim z→0 (−1)k (−k + 1 + z) · · · (z − 1)z ∞ 0 xzφ(k)(x)dx − 1 z φ(k−1)(0) (k − 1)! = lim z→0 (−1)k (−k + z + 1) · · · (z − 1) ∞ 0 xz − 1 z φ(k)(x)dx + 1 z (−1)k (−k + z + 1) · · · (z − 1) ∞ 0 φ(k)(x)dx − φ(k−1)(0) (k − 1)! = − 1 (k − 1)! ∞ 0 log x φ(k)(x)dx + lim z→0 ⎛ ⎝ φ(k−1)(0) z ⎡ ⎣ k−1 j=1 1 j − z − 1 (k − 1)! ⎤⎞ ⎦⎠ = − 1 (k − 1)! ∞ 0 log x φ(k)(x)dx + lim z→0 φ(k−1)(0) Ψk(z) − Ψk(0) z = − 1 (k − 1)! ∞ 0 log x φ(k)(x)dx + Ψ k (0) φ(k−1)(0) = − 1 (k − 1)! ∞ 0 log x φ(k)(x)dx + φ(k−1)(0) (k − 1)! k−1 j=1 1 j , where we have set Ψk(z) := k−1 j=1 1 j−z and used (1.10).
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