Digital Library of Mathematical Functions
The Gamma Function -- Integral Representations

Integral Representations

Contents

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