SOME INTEGRALS RELATED TO THE GAMMA
INTEGRAL
SVANTE JANSON
Abstract. We collect, for easy reference, some formulas related to the
Gamma integral.
We collect some formulas related to the Gamma integral. (None of the
formulas is new.) See also e.g. [1, Chapter 6] and [2, Section 5.9], where
further results are given. (Several of the formulas below appear in [2], but
we do not give individual references.)
All integrals are absolutely convergent unless we explicitly say otherwise.
We begin with the standard definition (Euler’s integral)
Z ∞
(1)
Γ(α) :=
xα−1 e−x dx,
Re α > 0.
0
I. Extensions to Re α < 0. For Re α < 0, the integral in (1) does not
converge, but if Re α ∈
/ Z we have the modifications
Z ∞
(2)
e−x − 1 xα−1 dx = Γ(α),
−1 < Re α < 0,
Z ∞ 0
(3)
e−x − 1 + x xα−1 dx = Γ(α),
−2 < Re α < −1,
0
and, in general, for any integer m > 0,
Z ∞
m
X
(−x)k α−1
−x
(4)
e −
x
dx = Γ(α),
k!
0
−m − 1 < Re α < −m.
k=0
Proof. Denote the integral in (4) by Iα,m . Then an integration by parts
gives
m
h
X
(−x)k α i∞
−x
(5) αIα,m = e −
x
+ Iα+1,m−1 = 0 + Iα+1,m−1 .
k!
0
k=0
For m = 0 we have −1 < Re α < 0 and then
Z ∞
Iα+1,−1 =
e−x xα dx = Γ(α + 1) = αΓ(α);
0
thus (4) for m = 0 follows from (5). (This is (2).) The general case now
follows by (5) and induction.
Date: 1 December, 2011; expanded 18 June, 2013.
1
2
SVANTE JANSON
Next we note the following extension of (2).
Z
(6)
∞
e−tx − 1 xα−1 dx = t−α Γ(α),
−1 < Re α < 0, Re t > 0.
0
Proof. For t > 0, this follows from (2) by a change of variables. The integral
in (6) converges for Re t > 0 and is a continuous function of t in this halfplane, analytic in the open half-plane Re t > 0. Hence the result follows by
analytic continuation.
II. sin and cos.
Z ∞
πα
xα−1 sin x dx = sin
(7)
Γ(α),
2
Z ∞ 0
πα
(8)
Γ(α),
xα−1 (1 − cos x) dx = − cos
2
0
−1 < Re α < 0,
−2 < Re α < 0,
Proof. For −1 < Re α < 0, these follow from (6) by taking t = ±i and using
Euler’s formulas. (Alternatively, for real α, by taking t = −i and taking
real and imaginary parts.) Then (8) extends to Re α > −2 by analytic
continuation.
In particular, taking α = −1 in (8) yields the wellknown
Z ∞
1 − cos x
π
(9)
dx = .
2
x
2
0
In fact, (7) extends to 0 6 Re α < 1, although the integral no longer is
absolutely convergent:
(10)
Z ∞
Z A
πα
xα−1 sin x dx := lim
xα−1 sin x dx = sin
Γ(α), −1 < Re α < 1.
A→∞
2
0
0
Proof. Integration by parts yields, using (8) and letting A → ∞,
Z A
Z A
α−1
A
α−1
x
sin x dx = x
(1 − cos x) 0 − (α − 1)
xα−2 (1 − cos x) dx
0
0
π(α − 1)
πα
→ 0 + (α − 1) cos
Γ(α − 1) = sin
Γ(α).
2
2
In particular, taking α = 0 in (10) yields the conditionally convergent
Z A
Z ∞
sin x
sin x
π
dx := lim
dx = .
(11)
A→∞ 0
x
x
2
0
There is also a corresponding conditionally convergent cosine integral,
related to (8):
(12)
Z ∞
Z A
πα
xα−1 cos x dx := lim
xα−1 cos x dx = cos
Γ(α), 0 < Re α < 1.
A→∞
2
0
0
SOME INTEGRALS RELATED TO THE GAMMA INTEGRAL
3
Proof. Integration by parts yields, using (7) and letting A → ∞,
Z A
Z A
α−1
A
α−1
xα−2 sin x dx
x
cos x dx = x
sin x 0 − (α − 1)
0
0
π(α − 1)
πα
→ 0 − (α − 1) sin
Γ(α − 1) = cos
Γ(α).
2
2
Another formula is:
Z ∞
(13)
e−ax − 1 + a sin x xα−1 dx
0
πα = a−α + a sin
Γ(α),
2
−2 < Re α < 0, Re a > 0.
Proof. If −1 < Re α < 0, this follows by (6) and (7). The general case
follows by analytic continuation.
In particular, taking α = −1 in (13) yields
Z ∞
(14)
e−ax − 1 + a sin x x−2 dx = a log a,
Re a > 0.
0
0
Proof. If f (a) := a−α + a sin πα
2 , then f (−1) = 0 and f (−1) = −a log a, and
if g(a) := 1/Γ(a) = a(a + 1)/Γ(a + 2), then g(−1) = 0 and g 0 (−1) = −1.
The result follows by l’Hôpital’s rule.
III. Subtracting on [0, 1] only.
Z 1
Z ∞
e−x xα−1 dx
(15)
e−x − 1 xα−1 dx +
0
1
Z ∞
=
e−x − 1{x < 1} xα−1 dx = Γ(α) − α−1 ,
−1 < Re α.
0
Proof. For Re α > 0, this follows from (1). The general case Re α > −1
follows by analytic continuation.
In particular, taking α = 0,
Z 1 −x
Z ∞ −x
Z ∞ −x
e −1
e
e − 1{x < 1}
(16)
dx +
dx =
dx = −γ.
x
x
x
0
1
0
Proof. As α → 0,
Γ(α + 1) − 1
→ Γ0 (1) = −γ.
α
We have also similar results with sin x and cos x in the integral.
Γ(α) − α−1 =
(17)
Z ∞
0
πα
1
xα−1 sin x−x1{x < 1} dx = sin
Γ(α)−
,
2
α+1
−3 < Re α < 1.
4
SVANTE JANSON
Here the integral is absolutely convergent if −3 < Re α < 0, and otherwise
conditionally convergent.
Proof. This follows from (7) when −1 < Re α < 0, and extends to −3 <
Re α < 0 by analytic continuation (with absolutely convergent integrals).
The case −1 < Re α < 1 follows similarly from (10).
In particular, taking α = 0, −1 and −2, cf. (11),
Z ∞
sin x − x1{x < 1}
π
(18)
dx = − 1,
x
2
Z0 ∞
sin x − x1{x < 1}
(19)
dx = 1 − γ,
x2
0
Z ∞
sin x − x1{x < 1}
π
dx = 1 − .
(20)
3
x
4
0
Proof. As ε → 0,
sin
sin
πε
1
π
Γ(ε) −
→ − 1,
2
ε+1
2
sin π(ε−1)
(ε − 1)−1 Γ(ε + 1) − 1
π(ε − 1)
1
2
Γ(ε − 1) − =
2
ε
ε
π(1−ε)
sin 2 (1 − ε)−1 Γ(ε + 1) − 1
=
ε
d
π(1 − ε)
→
sin
(1 − ε)−1 Γ(ε + 1) dε
2
ε=0
π
π
0
= − cos + 1 + Γ (1) = 1 − γ,
2
2
and
sin
− sin πε
1
1
π
π(ε − 2)
2 Γ(ε + 1)
Γ(ε − 2) −
=
+
→ − + 1.
2
ε−1
(ε − 2)(ε − 1)ε
1−ε
4
Similarly for cos x, with the integral absolutely convergent for −2 <
Re α < 0 and conditionally convergent for 0 6 Re α < 1:
(21)
Z ∞
πα
1
xα−1 cos x − 1{x < 1} dx = cos
Γ(α) − ,
−2 < Re α < 1.
2
α
0
Proof. This follows from (8) when −2 < Re α < 0. The case 0 < Re α < 1
follows directly from (12). The general case follows by integration by parts
SOME INTEGRALS RELATED TO THE GAMMA INTEGRAL
5
and (17), which yield
Z ∞
xα−1 cos x − 1{x < 1} dx
0
Z ∞
(α − 1)xα−2 sin x − x1{x < 1} − 1{x > 1} dx
=−
0
Z ∞
π(α − 1)
1
= −(α − 1) sin
(α − 1)xα−2 dx
Γ(α − 1) −
+
2
α
1
π(1 − α)
α−1
= sin
Γ(α) +
−1
2
α
1
πα
Γ(α) − .
= cos
2
α
In particular, taking α = 0 and −1, with the first integral conditionally
convergent and the second absolutely convergent,
Z ∞
cos x − 1{x < 1}
(22)
dx = −γ,
x
0
Z ∞
cos x − 1{x < 1}
π
(23)
dx = 1 − .
2
x
2
0
Proof. As ε → 0,
cos
cos πε
πε
1
2 Γ(ε + 1) − 1
Γ(ε) − =
2
ε
ε
d
πε
→
cos Γ(ε + 1) dε
2
ε=0
= Γ0 (1) = −γ
and
cos
sin πε
1
1
π
π(ε − 1)
2 Γ(ε + 1)
Γ(ε − 1) −
=
+
→ − + 1.
2
ε−1
(ε − 1)ε
1−ε
2
IV. Differences for different exponents.
(24)
Z ∞
e−ax −e−bx xα−1 dx = a−α −b−α Γ(α),
Re α > −1, Re a > 0, Re b > 0.
0
Proof. If Re α > 0 and a > 0, b > 0, this follows immediately from (1) by
separating the integral into two and changing variables. The case Re α > 0
now follows by analytic continuation in a and b, and this extends to Re α >
−1 by analytic continuation in α. (Cf. also (6).)
In particular, taking α = 0 we find:
Z ∞ −ax
e
− e−bx
b
(25)
dx = log b − log a = log ,
x
a
0
Re a, Re b > 0.
6
SVANTE JANSON
V. Another formula for γ.
Z ∞
Z ∞ −x
1
1 −x
e−x e
(26)
−
e
dx
=
−
dx = γ.
1 − e−x x
1 − e−x
x
0
0
Proof. We have, using (25),
Z ∞
Z ∞ −x
1 −x
e − 1 + x −x
1
−
e
dx
=
e dx
−x
1−e
x
x(1 − e−x )
0
0
Z ∞ −x
∞
e − 1 + x X −nx
e
dx
=
x
0
n=1
∞ Z ∞ −(n+1)x
X
e
− e−nx
=
+ e−nx dx
x
0
=
n=1
∞ X
N
log n − log(n + 1) +
n=1
X 1
1
= lim
− log(N + 1) = γ.
N →∞
n
n
n=1
VI. Other powers in the exponent. The change of variables x = y 1/β
yields immediately
Z ∞
1 α
β
,
Re α > 0.
(27)
xα−1 e−x dx = Γ
β
β
0
and, in particular,
Z ∞
β
(28)
e−x dx = Γ(1 + 1/β),
β > 0.
0
References
[1] M. Abramowitz & I. A. Stegun, eds., Handbook of Mathematical Functions. 9th printing. Dover, New York, 1972. Also
available at http://people.maths.ox.ac.uk/~macdonald/aands/
abramowitz_and_stegun.pdf
[2] Frank W. J. Olver, Daniel W. Lozier, Ronald F. Boisvert and Charles
W. Clark, NIST Handbook of Mathematical Functions. Cambridge Univ.
Press, 2010.
Also available as NIST Digital Library of Mathematical Functions,
http://dlmf.nist.gov/
Department of Mathematics, Uppsala University, PO Box 480, SE-751 06
Uppsala, Sweden
E-mail address: [email protected]
URL: http://www2.math.uu.se/∼svante/