The gamma and the beta function

As mentioned in the book [1], see page 6, the integral representation (1.1.18) is often takenas a definition for the gamma function (z). The advantage of this alternative definition isthat we might avoid the use of infinite products (see appendix A).

Definition 1.

(z) =

0

ettz1 dt, Re z > 0. (1)

From this definition it is clear that (z) is analytic for Re z > 0. By using integration byparts we find that

(z + 1) =

0

ettz dt = 0

tz det = ettz0

+

0

et dtz

= z

0

ettz1 dt = z(z), Re z > 0.

Hence we have

Theorem 1.(z + 1) = z(z), Re z > 0. (2)

Further we have

(1) =

0

et dt = et0

= 1. (3)

Combining (2) and (3), this leads to

(n+ 1) = n!, n = 0, 1, 2, . . . . (4)

The functional relation (2) can be used to find an analytic continuation of the gammafunction for Re z 0. For Re z > 0 the gamma function (z) is defined by (1). The functionalrelation (2) also holds for Re z > 0.

Let 1 < Re z 0, then we have Re (z+1) > 0. Hence, (z+1) is defined by the integralrepresentation (1). Now we define

(z) =(z + 1)

z, 1 < Re z 0, z 6= 0.

Then the gamma function (z) is analytic for Re z > 1 except z = 0. For z = 0 we havelimz0

z(z) = limz0

(z + 1) = (1) = 1.

This implies that (z) has a single pole at z = 0 with residue 1.This process can be repeated for 2 < Re z 1, 3 < Re z 2, etcetera. Then

the gamma function turns out to be an analytic function on C except for single poles atz = 0,1,2, . . .. The residue at z = n equals

limzn(z + n)(z) = limzn(z + n)

(z + 1)

z= lim

zn(z + n)1

z

1

z + 1 1

z + n 1(z + n+ 1)

z + n

=(1)

(n)(n+ 1) (1) =(1)nn!

, n = 0, 1, 2, . . . .

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As indicated in the book [1], see page 8, the limit formula (1.1.5) can be obtained fromthe integral representation (1) by using induction as follows. We first prove that 1

0(1 t)ntz1 dt = n!

(z)n+1(5)

for Re z > 0 and n = 0, 1, 2, . . .. Here the shifted factorial (a)k is defined by

Definition 2.

(a)k = a(a+ 1) (a+ k 1), k = 1, 2, 3, . . . and (a)0 = 1. (6)In order to prove (5) by induction we first take n = 0 to obtain for Re z > 0 1

0tz1 dt =

tz

z

10

=1

z=

0!

(z)1.

Now we assume that (5) holds for n = k. Then we have 10

(1 t)k+1tz1 dt = 10

(1 t)(1 t)ktz1 dt = 10

(1 t)ktz1 dt 10

(1 t)ktz dt

=k!

(z)k+1 k!

(z + 1)k+1=

k!

(z)k+2(z + k + 1 z) = (k + 1)!

(z)k+2,

which is (5) for n = k + 1. This proves that (5) holds for all n = 0, 1, 2, . . ..Now we set t = u/n into (5) to find that

1

nz

n0

(1 u

n

)nuz1 du =

n!

(z)n+1=

n0

(1 u

n

)nuz1 du =

n!nz

(z)n+1.

Since we havelimn

(1 u

n

)n= eu,

we conclude that

(z) =

0

euuz1 du = limn

n!nz

(z)n+1.

The beta function B(u, v) is also defined by means of an integral:

Definition 3.

B(u, v) =

10tu1(1 t)v1 dt, Reu > 0, Re v > 0. (7)

This integral is often called the beta integral. From the definition we easily obtain thesymmetry

B(u, v) = B(v, u), (8)

since we have by using the substitution t = 1 s

B(u, v) =

10tu1(1 t)v1 dt =

01

(1 s)u1sv1 ds = 10sv1(1 s)u1 ds = B(v, u).

The connection between the beta function and the gamma function is given by the fol-lowing theorem:

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Theorem 2.

B(u, v) =(u)(v)

(u+ v), Reu > 0, Re v > 0. (9)

In order to prove this theorem we use the definition (1) to obtain

(u)(v) =

0

ettu1 dt 0

essv1 ds = 0

0

e(t+s)tu1sv1 dt ds.

Now we apply the change of variables t = xy and s = x(1 y) to this double integral. Notethat t + s = x and that 0 < t < and 0 < s < imply that 0 < x < and 0 < y < 1.The Jacobian of this transformation is

(t, s)

(x, y)=

y x1 y x = xy x+ xy = x.

Since x > 0 we conclude that dt ds =

(t, s)(x, y) dx dy = x dx dy. Hence we have

(u)(v) =

10

0

exxu1yu1xv1(1 y)v1x dx dy

=

0

exxu+v1 dx 10yu1(1 y)v1 dy = (u+ v)B(u, v).

This proves (9).There exist many useful forms of the beta integral which can be obtained by an appropriate

change of variables. For instance, if we set t = s/(s+ 1) into (7) we obtain

B(u, v) =

10tu1(1 t)v1 dt =

0

su1(s+ 1)u+1(s+ 1)v+1(s+ 1)2 ds

=

0

su1

(s+ 1)u+vds, Reu > 0, Re v > 0.

This proves

Theorem 3.

B(u, v) =

0

su1

(s+ 1)u+vds, Reu > 0, Re v > 0. (10)

If we set t = cos2 into (7) we find that

B(u, v) =

10tu1(1 t)v1 dt = 2

0pi/2

(cos)2u2(sin)2v2 cos sind

= 2

pi/20

(cos)2u1(sin)2v1 d, Reu > 0, Re v > 0.

Hence we have

Theorem 4.

B(u, v) = 2

pi/20

(cos)2u1(sin)2v1 d, Reu > 0, Re v > 0. (11)

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Finally, the substitution t = (s a)/(b a) in (7) leads to

B(u, v) =

10tu1(1 t)v1 dt

=

ba

(s a)u1(b a)u+1(b s)v1(b a)v+1(b a)1 ds

= (b a)uv+1 ba

(s a)u1(b s)v1 ds, Reu > 0, Re v > 0.

Hence we have

Theorem 5. ba

(s a)u1(b s)v1 ds = (b a)u+v1B(u, v), Reu > 0, Re v > 0. (12)

The special case a = 1 and b = 1 is of special interest as we will see later: 11

(1 + s)u1(1 s)v1 ds = 2u+v1B(u, v), Reu > 0, Re v > 0.

The different forms for the beta function have a lot of consequences. For instance, if weset u = v = 1/2 in (9) we find that

B(1/2, 1/2) =(1/2)(1/2)

(1)= {(1/2)}2 .

On the other hand, we have by using (11)

B(1/2, 1/2) = 2

pi/20

d = 2 pi2

= pi.

This implies that(1/2) =

pi. (13)

By using the transformation x2 = t we now easily obtain the value of the normal integral

ex2dx = 2

0

ex2dx =

0

ett1/2 dt = (1/2) =pi. (14)

The combination of (9) and (11) can be used to compute integrals such as pi/20

(cos)5(sin)7 d =1

2B(3, 4) = 1

2 (3)(4)

(7)=

1

2 2! 3!

6!=

1

2 2

4 5 6 =1

120,

pi/20

(cos)7(sin)4 d =1

2B(4, 5/2) = 1

2 (4)(5/2)

(13/2)=

1

2 3!52 72 92 112

=1

2 6 2

4

5 7 9 11 =16

1155

4

and pi/20

(cos)4(sin)6 d =1

2B(5/2, 7/2) = 1

2 (5/2)(7/2)

(6)

=1

232 12 (1/2) 52 32 12 (1/2)

5!=

5 32 pi26 2 3 4 5 =

3pi

29=

3pi

512.

Another important consequence of (9) and (11) is Legendres duplication formula for thegamma function:

Theorem 6.(z)(z + 1/2) = 212z

pi (2z), Re z > 0. (15)

In order to prove this we use (11) and the transformation 2 = to find that

B(z, z) = 2

pi/20

(cos)2z1(sin)2z1 d = 2 212z pi/20

(sin 2)2z1 d

= 212z pi0

(sin )2z1 d = 212z 2 pi/20

(sin )2z1 d = 212z B(z, 1/2).

Now we apply (9) to obtain

(z)(z)

(2z)= B(z, z) = 212z B(z, 1/2) = 212z (z)(1/2)

(z + 1/2), Re z > 0.

Finally, by using (13), this implies that

(z)(z + 1/2) = 212zpi (2z), Re z > 0.

This proves the theorem.Legendres duplication formula can be generalized to Gausss multiplication formula:

Theorem 7.

(z)

n1k=1

(z + k/n) = n1/2nz(2pi)(n1)/2(nz), n {1, 2, 3, . . .}. (16)

The case n = 1 is trivial and the case n = 2 is Legendres duplication formula.Another property of the gamma function is given by Eulers reflection formula:

Theorem 8.(z)(1 z) = pi

sinpiz, z 6= 0,1,2, . . . . (17)

This can be shown by using contour integration in the complex plane as follows. First werestrict to real values of z, say z = x with 0 < x < 1. By using (9) and (10) we have

(x)(1 x) = B(x, 1 x) = 0

tx1

t+ 1dt.

In order to compute this integral we consider the contour integralCzx1

1 z dz,

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where the contour C consists of two circles about the origin of radii R and respectively,which are joined along the negative real axis from R to . Move along the outer circlewith radius R in the positive (counterclockwise) direction and along the inner circle withradius in the negative (clockwise) direction. By the residue theorem we have

Czx1

1 z dz = 2pii,

when zx1 has its principal value. This implies that

2pii =C1

zx1

1 z dz +C2

zx1

1 z dz +C3

zx1

1 z dz +C4

zx1

1 z dz,

where C1 denotes the outer circle with radius R, C2 denotes the line segment from R to ,C3 denotes the inner circle with radius and C4 denotes the line segment from to R.Then we have by writing z = Rei for the outer circle

C1

zx1

1 z dz = pipi

Rx1ei(x1)

1Rei d(Rei

)=

pipi

iRxeix

1Rei d.

In the same way we have by writing z = ei for the inner circleC3

zx1

1 z dz = pipi

ixeix

1 ei d.

For the line segment from R to we have by writing z = t = tepiiC2

zx1

1 z dz = R

tx1ei(x1)pi

1 + td(tepii

)=

R

tx1eixpi

1 + tdt.

In the same way we have by writing z = t = tepiiC4

zx1

1 z dz = R

tx1eixpi

1 + tdt.

Since 0 < x < 1 we have

limR

pipi

iRxeix

1Rei d = 0 and lim0 pipi

ixeix

1 ei d = 0.

Hence we have

2pii = 0tx1eixpi

1 + tdt+

0

tx1eixpi

1 + tdt,

or

2pii = (eixpi eixpi) 0

tx1

1 + tdt =

0

tx1

1 + tdt =

2pii

eixpi eixpi =pi

sinpix.

This proves the theorem for real values of z, say z = x with 0 < x < 1. The full result followsby analytic continuation. Alternatively, the result can be obtained as follows. If (17) holdsfor real values of z with 0 < z < 1, then it holds for all complex z with 0 < Re z < 1 by

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analyticity. Then it also holds for Re z = 0 with z 6= 0 by continuity. Finally, the full resultfollows for z shifted by integers using (2) and sin(z + pi) = sin z. Note that (17) holds forall complex values of z with z 6= 0,1,2, . . .. Instead of (17) we may write

1

(z)(1 z) =sinpiz

pi, (18)

which holds for all z C.Now we will prove an asymptotic formula which is due to Stirling. First we define

Definition 4. Two functions f and g of a real variable x are called asymptotically equal,notation

f g for x, if limx

f(x)

g(x)= 1.

Now we have Stirlings formula:

Theorem 9.(x+ 1) xx+1/2ex

2pi, x. (19)

Here x denotes a real variable. This can be proved as follows. Consider

(x+ 1) =

0

ettx dt,

where x R. Then we obtain by using the transformation t = x(1 + u)

(x+ 1) =

1

ex(1+u)xx(1 + u)xx du = xx+1ex 1

exu(1 + u)x du

= xx+1ex 1

ex(u+ln(1+u)) du.

The function f(u) = u + ln(1 + u) equals zero for u = 0. For other values of u we havef(u) < 0. This implies that the integrand of the last integral equals 1 at u = 0 and that thisintegrand becomes very small for large values of x at other values of u. So for large values ofx we only have to deal with the integrand near u = 0. Note that we have

f(u) = u+ ln(1 + u) = 12u2 +O(u3) for u 0.

This implies that 1

ex(u+ln(1+u)) du

exu2/2 du for x.

If we set u = t

2/x we have by using the normal integral (14)

exu2/2 du = x1/2

2

et2dt = x1/2

2pi.

Hence we have(x+ 1) xx+1/2ex

2pi, x,

which proves the theorem.

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Note that Stirlings formula implies that

n! nnen

2pin for nand that

(n+ a)

(n+ b) nab for n.

The theorem can be extended for z in the complex plane:

Theorem 10. For > 0 we have

(z + 1) zz+1/2ez

2pi for |z| with | arg z| pi . (20)

Stirlings asymptotic formula can be used to give an alternative proof for Eulers reflectionformula (17) for the gamma function. Consider the function

f(z) = (z)(1 z) sinpiz.Then we have

f(z + 1) = (z + 1)(z) sinpi(z + 1) = z(z) (1 z)z sinpiz = f(z).

Hence, f is periodic with period 1. Further we have

limz0

f(z) = limz0

(z)(1 z) sinpiz = limz0

(z + 1)(1 z)sinpizz

= pi, (21)

which implies that f has no poles. Hence, f is analytic and periodic with period 1. Now wewant to apply Liouvilles theorem for entire functions, id est functions which are analytic onthe whole complex plane:

Theorem 11. Every bounded entire function is constant.

Therefore, we want to show that f is bounded. Since f is periodic with period 1 we consider0 Re z 1, say z = x+ iy with x and y real and 0 x 1. Then we have

sinpiz =eipiz eipiz

2i 1

2ieipiz =

i

2eipiz for y .

Now we apply Stirlings formula to obtain

f(z) = (z)(1 z) sinpiz zz1/2ez

2pi (z)z+1/2ez

2pii

2eipiz.

For y > 0 we have z/z = epii. Hence, f(z) pi for y . This implies that f is bounded.So, Liouvilles theorem implies that f is constant. By using (21) we conclude that f(z) = pi,which proves Eulers reflection formula (17) or (18).

Stirlings formula can also be used to give an alternative proof for Legendres duplicationformula (15). Consider the function

g(z) = 22z1(z)(z + 1/2)

(1/2)(2z).

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Then we have by using (2)

g(z + 1) = 22z+1(z + 1)(z + 3/2)

(1/2)(2z + 2)= 22z+1

z(z)(z + 1/2)(z + 1/2)

(1/2)(2z + 1)2z(2z)= g(z).

Further we have by using (13) and Stirlings asymptotic formula (20)

g(z) 22z1 zz1/2ez

2pi zzez

2pi

pi 22z1/2z2z1/2e2z

2pi= 1.

This implies thatlimn g(z + n) = 1,

also for integer values for n. On the other hand we have g(z + n) = g(z) for integer values ofn. This implies that g(z) = 1 for all z. This proves (15).

Finally we have the digamma function (z) which is related to the gamma function. Thisfunction (z) is defined as follows.

Definition 5.

(z) =(z)(z)

=d

dzln (z), z 6= 0,1,2, . . . . (22)

A property of this digamma function that is easily proved by using (2) is given by the followingtheorem:

Theorem 12.

(z + 1) = (z) +1

z. (23)

By using (2) we have

(z + 1) =d

dzln (z + 1) =

d

dzln (z(z)) =

d

dzln z +

d

dzln (z) =

1

z+ (z).

This proves the theorem. Iteration of (23) easily leads to

Theorem 13.

(z + n) = (z) +1

z+

1

z + 1+ . . .+

1

z + n 1 , n = 1, 2, 3, . . . . (24)Another property of the digamma function is given by

Theorem 14.(z) (1 z) = pi

tanpiz, z 6= 0,1,2, . . . . (25)

The proof of this theorem is based on (17). We have

(z) (1 z) = ddz

ln (z) +d

dzln (1 z) = d

dzln ((z)(1 z))

=d

dzln

pi

sinpiz=

sinpiz

pi pi

2 cospiz

(sinpiz)2= pi

tanpiz.

References

[1] G.E. Andrews, R. Askey and R. Roy, Special Functions. Encyclopedia of Mathemat-ics and its Applications 71, Cambridge University Press, 1999.

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