Integral Representations
Contents
- Gamma Function
- Psi Function and Euler's Constant
Gamma Function
1
μ
Γ
(
ν
μ
)
1
z
ν
μ
=
∫
0
∞
exp
(
-
z
t
μ
)
t
ν
-
1
ⅆ
t
ℜ
ν
>
0
,
μ
>
0
, and
ℜ
z
>
0
. (The fractional powers have their principal values.)
Hankel's Loop Integral
1
Γ
(
z
)
=
1
2
π
ⅈ
∫
-
∞
(
0
+
)
ⅇ
t
t
-
z
ⅆ
t
where the contour begins at
-
∞
, circles the origin once in the positive direction, and returns to
-
∞
.
t
-
z
has its principal value where
t
crosses the positive real axis, and is continuous.
t
-plane. Contour for Hankel's loop integral.
c
-
z
Γ
(
z
)
=
∫
-
∞
∞
|
t
|
2
z
-
1
ⅇ
-
c
t
2
ⅆ
t
c
>
0
,
ℜ
z
>
0
where the path is the real axis.
Γ
(
z
)
=
∫
1
∞
t
z
-
1
ⅇ
-
t
ⅆ
t
+
∑
k
=
0
∞
(
-
1
)
k
(
z
+
k
)
k
!
z
≠
0
,
-
1
,
-
2
,
…
Γ
(
z
)
=
∫
0
∞
t
z
-
1
(
ⅇ
-
t
-
∑
k
=
0
n
(
-
1
)
k
t
k
k
!
)
ⅆ
t
-
n
-
1
<
ℜ
z
<
-
n
Γ
(
z
)
cos
(
1
2
π
z
)
=
∫
0
∞
t
z
-
1
cos
t
ⅆ
t
0
<
ℜ
z
<
1
,
Γ
(
z
)
sin
(
1
2
π
z
)
=
∫
0
∞
t
z
-
1
sin
t
ⅆ
t
-
1
<
ℜ
z
<
1
.
Γ
(
1
+
1
n
)
cos
(
π
2
n
)
=
∫
0
∞
cos
(
t
n
)
ⅆ
t
n
=
2
,
3
,
4
,
…
Γ
(
1
+
1
n
)
sin
(
π
2
n
)
=
∫
0
∞
sin
(
t
n
)
ⅆ
t
n
=
2
,
3
,
4
,
…
.
Binet's Formula
ln
Γ
(
z
)
=
(
z
-
1
2
)
ln
z
-
z
+
1
2
ln
(
2
π
)
+
2
∫
0
∞
arctan
(
t
z
)
ⅇ
2
π
t
-
1
ⅆ
t
where
|
ph
z
|
<
π
2
and the inverse tangent has its principal value.
ln
Γ
(
z
+
1
)
=
-
γ
z
-
1
2
π
ⅈ
∫
-
c
-
∞
ⅈ
-
c
+
∞
ⅈ
π
z
-
s
s
sin
(
π
s
)
ζ
(
-
s
)
ⅆ
s
where
|
ph
z
|
≤
π
-
δ
(
<
π
),
1
<
c
<
2
, and
ζ
(
s
)
For additional representations see
Whittaker and Watson(1927)
Psi Function and Euler's Constant
For
ℜ
z
>
0
,
ψ
(
z
)
=
∫
0
∞
(
ⅇ
-
t
t
-
ⅇ
-
z
t
1
-
ⅇ
-
t
)
ⅆ
t
ψ
(
z
)
=
ln
z
+
∫
0
∞
(
1
t
-
1
1
-
ⅇ
-
t
)
ⅇ
-
t
z
ⅆ
t
ψ
(
z
)
=
∫
0
∞
(
ⅇ
-
t
-
1
(
1
+
t
)
z
)
ⅆ
t
t
ψ
(
z
)
=
ln
z
-
1
2
z
-
2
∫
0
∞
t
ⅆ
t
(
t
2
+
z
2
)
(
ⅇ
2
π
t
-
1
)
ψ
(
z
)
+
γ
=
∫
0
∞
ⅇ
-
t
-
ⅇ
-
z
t
1
-
ⅇ
-
t
ⅆ
t
=
∫
0
1
1
-
t
z
-
1
1
-
t
ⅆ
t
ψ
(
z
+
1
)
=
-
γ
+
1
2
π
ⅈ
∫
-
c
-
∞
ⅈ
-
c
+
∞
ⅈ
π
z
-
s
-
1
sin
(
π
s
)
ζ
(
-
s
)
ⅆ
s
where
|
ph
z
|
≤
π
-
δ
(
<
π
)
and
1
<
c
<
2
.
γ
=
-
∫
0
∞
ⅇ
-
t
ln
t
ⅆ
t
=
∫
0
∞
(
1
1
+
t
-
ⅇ
-
t
)
ⅆ
t
t
=
∫
0
1
(
1
-
ⅇ
-
t
)
ⅆ
t
t
-
∫
1
∞
ⅇ
-
t
ⅆ
t
t
=
∫
0
∞
(
ⅇ
-
t
1
-
ⅇ
-
t
-
ⅇ
-
t
t
)
ⅆ
t