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
˜
F
−1+z
(φ) =
1
z

0
xzφ
(x)dx.
Since
Resz=0
˜
F
−1+z
(φ) = φ(0),
we have
lim
z→0
˜
F
−1+z
(φ)
1
z
Resz=0
˜
F
−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
˜
F −k+z(φ) =
(−1)k
(−k + 1 + z) · · · (z 1)z

0
xzφ(k)(x)dx.
Since Resz=0
˜
F −k+z(φ) =
φ(k−1)(0)
(k−1)!
, we have
lim
z→0
˜
F −k+z(φ)
1
z
Resz=0
˜
F −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
+ Ψk(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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