# 2 Deﬁnitions of the gamma function

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## Introduction to the Gamma Function

### numbers.computation.free.fr/Constants/constants.html February 4, 2002

Abstract

An elementary introduction to the celebrated gamma function Γ(x) and its various representations. Some of its most important properties are described.

### 1Introduction

The gamma function was ﬁrst introduced by the Swiss mathematician Leon- hard Euler (1707-1783) in his goal to generalize the factorial to non integer values. Later, because of its great importance, it was studied by other eminent mathematicians like Adrien-Marie Legendre (1752-1833), Carl Friedrich Gauss (1777-1855), Christoph Gudermann (1798-1852), Joseph Liouville (1809-1882), Karl Weierstrass (1815-1897), Charles Hermite (1822-1901), ... as well as many others.

The gamma function belongs to the category of the special transcendental functions and we will see that some famous mathematical constants are occur- ring in its study.

It also appears in various area as asymptotic series, deﬁnite integration, hypergeometric series, Riemann zeta function, number theory ...

Some of the historical background is due to Godefroy’s beautiful essay on this function [9] and the more modern textbook [3] is a complete study.

### 2.1Deﬁnite integral

During the years 1729 and 1730 ([9], [12]), Euler introduced an analytic function which has the property to interpolate the factorial whenever the argument of the function is an integer. In a letter from January 8, 1730 to Christian Goldbach he proposed the following deﬁnition :

Deﬁnition 1 (Euler, 1730) Let x > 0

Γ(x) =

 1

0 (− log(t))x−1dt. (1)

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By elementary changes of variables this historical deﬁnition takes the more usual forms :

Theorem 2 For x > 0

Γ(x) =



0 tx−1e−tdt, (2)

or sometimes

Γ(x) = 2



0 t2x−1e−t2dt. (3)

Proof. Use respectively the changes of variable u = − log(t) and u2 =

− log(t) in (1).

From this theorem, we see that the gamma function Γ(x) (or the Eulerian integral of the second kind ) is well deﬁned and analytic for x > 0 (and more generally for complex numbers x with positive real part).

The notation Γ(x) is due to Legendre in 1809 [11] while Gauss expressed it by Π(x) (which represents Γ(x + 1)).

The derivatives can be deduced by diﬀerentiating under the integral sign of (2)

Γ(x) =



0 tx−1e−tlog(t)dt, Γ(n)(x) =



0 tx−1e−tlogn(t)dt.

2.1.1 Functional equation We have obviously

Γ(1) =



0 e−tdt = 1 (4)

and for x > 0, an integration by parts yields Γ(x + 1) =



0 txe−tdt = [−txe−t]0 + x



0 tx−1e−tdt = xΓ(x), (5) and the relation Γ(x + 1) = xΓ(x) is the important functional equation.

For integer values the functional equation becomes Γ(n + 1) = n!,

and it’s why the gamma function can be seen as an extension of the factorial function to real non null positive numbers.

A natural question is to determine if the gamma function is the only solution of the functional equation ? The answer is clearly no as may be seen if we consider, for example, the functions cos(2mπx)Γ(x), where m is any non null integer and which satisfy both (4) and (5). But the following result states that under an additional condition the gamma function is the only solution of this equation.

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Theorem 3 (Bohr-Mollerup, 1922, [6]) There is aunique function f : ]0, +∞[→

]0, +∞[ such as log(f (x)) is convex and f (1) = 1, f (x + 1) = xf (x).

Proof. An elementary one is given in [2].

Other conditions may also work as well, see again [2].

It’s also possible to extend this function to negative values by inverting the functional equation (which becomes a deﬁnition identity for−1 < x < 0)

Γ(x) = Γ(x + 1)

x ,

and for example Γ(−1/2) = −2Γ(1/2). Reiteration of this identity allows to deﬁne the gamma function on the whole real axis except on the negative integers (0, −1, −2, ...). For any non null integer n, we have

Γ(x) = Γ(x + n)

x(x + 1)...(x + n − 1) x + n > 0. (6) Suppose that x = −n + h with h being small, then

Γ(x) = Γ(1 + h)

h(h − 1)...(h − n) (−1)n

n!h when h → 0,

so Γ(x) possesses simple poles at the negative integers −n with residue (−1)n/n!

(see the plot of the function 2.1.1).

In fact, also by mean of relation (6), the gamma function can be deﬁned in the whole complex plane.

–4 –3 –2 –1 0 1 2 3 4

–4 –3 –2 –1 1 2 3 4

x

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Γ(x) function

### 2.2Another deﬁnition by Euler and Gauss

In another letter written in October 13, 1729 also to his friend Goldbach, Euler gave another equivalent deﬁnition for Γ(x).

Deﬁnition 4 (Euler, 1729 and Gauss, 1811) Let x > 0 and deﬁne

Γp(x) = p!px

x(x + 1)...(x + p) = px

x(1 + x/1)...(1 + x/p), (7) then

Γ(x) = lim

p→∞ Γp(x). (8)

(Check the existence of this limit). This approach, using an inﬁnite product, was also chosen, in 1811, by Gauss in his study of the gamma function [8].

Clearly

Γp(1) = p!

1(1 + 1)...(1 + p)p = p p + 1, and Γp(x + 1) = p!px+1

(x + 1)...(x + p + 1) = p

x + p + 1xΓp(x), hence

Γ(1) = 1, Γ(x + 1) = xΓ(x).

We retrieve the functional equation veriﬁed by Γ(x).

It’s interesting to observe that the deﬁnition is still valid for negative values of x, except on the poles (0, −1, −2, ...). Using this formulation is often more convenient to establish new properties of the gamma function.

### 2.3Weierstrass formula

The relation

px= ex log(p)= ex(log(p)−1−1/2−...−1/p)ex+x/2+...+x/p, entails

Γp(x) = 1 x

1 x + 1

2

x + 2... p

x + ppx=ex(log(p)−1−1/2−...−1/p)ex+x/2+...+x/p

x(1 + x)(1 + x/2)...(1 + x/p) , Γp(x) = ex(log(p)−1−1/2−...−1/p)1

x ex 1 + x

ex/2

1 + x/2· · · ex/p 1 + x/p.

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Now Euler’s constant is deﬁned by γ = limp→∞

 1 + 1

2+ ... + 1

p− log(p)



= 0.5772156649015328606..., and therefore follows the Weierstrass form of the gamma function.

Theorem 5 (Weierstrass) For any real number x, except on the negative inte- gers (0, −1, −2, ...), we have the inﬁnite product

1

Γ(x) = xeγx

 p=1

 1 +x

p



e−x/p. (9)

From this product we see that Euler’s constant is deeply related to the gamma function and the poles are clearly the negative or null integers. Ac- cording to Godefroy [9], Euler’s constant plays in the gamma function theory a similar role as π in the circular functions theory.

It’s possible to show that Weierstrass form is also valid for complex numbers.

### 3Some special values of Γ(x)

Except for the integer values of x = n for which Γ(n) = (n − 1)!

some non integers values have a closed form.

The change of variable t = u2 gives Γ(1/2) =



0

e√−t tdt = 2



0 e−u2du = 2

√π 2 =

π.

The functional equation (5) entails for positive integers n (see [1]) Γ

 n +1

2



=1.3.5...(2n − 1) 2n

√π, (10)

Γ

 n +1

3



=1.4.7...(3n − 2)

3n Γ

1 3

 ,

Γ

 n +1

4



=1.5.9...(4n − 3)

4n Γ

1 4

 , and for negative values

Γ



−n +1 2



= (−1)n2n 1.3.5...(2n − 1)

√π.

No basic expression is known for Γ(1/3) or Γ(1/4), but it was proved that those numbers are transcendental (respectively by Le Lionnais in 1983 and Chudnovsky in 1984).

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Up to 50 digits, the numerical values of some of those constants are :





Γ (1/2) = 1.77245385090551602729816748334114518279754945612238...

Γ (1/3) = 2.67893853470774763365569294097467764412868937795730...

Γ (1/4) = 3.62560990822190831193068515586767200299516768288006...

Γ (1/5) = 4.59084371199880305320475827592915200343410999829340...

.

For example, thanks to the very fast converging formula (which is based on the expression (34) and uses the Arithmetic-Geometric Mean AGM, [7] )

Γ2(1/4) = (2π)3/2 AGM (√

2, 1),

this constant was computed to more than 50 millions digits by P. Sebah and M.

Tommila [10]. Similar formulae are available for other fractional arguments like Γ (1/3) ...

### 4.1The complement formula

There is an important identity connecting the gamma function at the comple- mentary values x and 1 − x. One way to obtain it is to start with Weierstrass formula (9) which yields

1 Γ(x)

1

Γ(−x) =−x2eγxe−γx

 p=1



1 +x p

 e−x/p

 1−x

p

 ex/p

 .

But the functional equation gives Γ(−x) = −Γ(1 − x)/x and the equality writes as

1

Γ(x)Γ(1 − x) = x

 p=1

 1−x2

p2

 ,

and using the well-known inﬁnite product :

sin(π x) = π x

 p=1

 1−x2

p2



ﬁnally gives

Γ(x)Γ(1 − x) = π

sin πx. (11)

Relation (11) is the complement (or reﬂection) formula and is valid when x and 1− x are not negative or null integers and it was discovered by Euler.

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For example, if we apply this formula for the values x = 1/2, x = 1/3, x = 1/4 we ﬁnd

Γ

1 2



= π,

Γ

1 3

 Γ

2 3



=2π√ 3 3 , Γ

1 4

 Γ

3 4



= π√ 2.

### 4.2Duplication and Multiplication formula

In 1809, Legendre obtained the following duplication formula [11].

Theorem 6 (Legendre, 1809)

Γ(x)Γ(x + 1/2) =

√π

22x−1Γ(2x). (12)

Proof. Hint : an easy proof can lie on the expression of Γp(x) and Γp(x+1/2) from (7), then make the product and ﬁnd the limit as p → ∞.

Notice that by applying the duplication formula for x = 1/2, we retrieve the value of Γ(1/2), while x = 1/6 permits to compute

Γ

1 6



= 2−1/3

3 πΓ2

1 3

 .

This theorem is the special case when n = 2 of the more general result known as Gauss multiplication formula :

Theorem 7 (Gauss)

Γ (x) Γ

 x + 1

n

 Γ

 x + 2

n

 ...Γ



x + n − 1 n



= (2π)(n−1)/2n1/2−nxΓ (nx) Proof. Left as exercise.

Corollary 8 (Euler)

Γ

1 n

 Γ

2 n

 ...Γ

n − 1 n



= (2π)√(n−1)/2 n Proof. Set x = 1/n in the Gauss multiplication formula.

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### 4.3Stirling’s formula

It’s of interest to study how the gamma function behaves when the argument x becomes large. If we restrict the argument x to integral values n, the following result, due to James Stirling (1692-1730) and Abraham de Moivre (1667-1754) is famous and of great importance :

Theorem 9 (Stirling-De Moivre, 1730) If the integer n tends to inﬁnite we have the asymptotic formula

Γ(n + 1) = n! ∼√

2πnnne−n. (13)

Proof. See [2] for a complete proof. You may obtain a weaker approxima- tion by observing that the area under the curve log(x) with x ∈ [1, n] is well approximated by the trapezoidal rule, therefore

 n

1 log(x)dx = n log(n) − n + 1

=

n−1

k=1

log(k) + log(k + 1)

2 + Rn= log((n − 1)!) + 1

2log(n) + Rn

and because Rn = O(1) (check this!), we ﬁnd log(n!) ≈ n log(n) + 1

2log(n) − n + C which gives this weaker result

n! ≈ eCnn ne−n.

Stirling’s formula is remarkable because the pure arithmetic factorial func- tion is equivalent to an expression containing important analytic constants like (

2, π, e).

There is an elementary way to improve the convergence of Stirling’s formula.

Suppose you can write n! =√

2πnnne−n

1 +a1

n +a2

n2 + ...

, then this relation is still valid for n + 1

(n + 1)! =

2π(n + 1)(n + 1)(n+1)e−(n+1)



1 + a1

n + 1+ a2

(n + 1)2 + ...

 (14)

but we also have (n + 1)! = (n + 1)n! giving (n + 1)! = (n + 1)√

2πnnne−n

1 + a1

n +a2

n2 + ...

. (15)

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We now compare relations (14) and (15) when n becomes large. This gives after some simpliﬁcations and classical series expansions

1 + a1

n + a2

n2 + ...

=

 1 + 1

n

n+1/2 e−1



1 + a1

n + 1+ a2

(n + 1)2 + ...



= 1 + a1

n +a2− a1+121

n2 +

1312a1− 2a2+ a3121

n3 + ...

and after the identiﬁcation comes

−a1+ 1 12 = 0, 13

12a1− 2a2 1 12 = 0,

...

Therefore we found, by elementary means, the ﬁrst correcting terms of the formula to be : a1= 1/12, a2= 1/288, ... A more eﬃcient (but less elementary) way to ﬁnd more terms is to use the Euler-Maclaurin asymptotic formula.

In fact the following theorem is a generalization of Stirling’s formula valid for any real number x :

Theorem 10 When x → ∞, we have the famous Stirling’s asymptotic formula [1]

Γ(x + 1) =√

2πxxxe−x

 1 + 1

12x+ 1

288x2 139

51840x3 571 2488320x4...

 . (16)

For example here are some approximations of the factorial using diﬀerent values for n :

n n! Stirling formula + correction 1/ (12n)

5 120 118 119

10 3628800 3598695 3628684

20 2432902008176640000 2422786846761133393 2432881791955971449

### 5Series expansion

To estimate the gamma function near a point it’s possible to use some series expansions at this point. Before doing this we need to introduce a new function which is related to the derivative of the gamma function.

### 5.1The digamma and polygamma functions

Many of the series involving the gamma function and its derivatives may be derived from the Weierstrass formula. By taking the logarithm on both sides of

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the Weierstrass formula (9) we ﬁnd the basic relation

− log(Γ(x)) = log(x) + γx + p=1

 log

 1 +x

p



−x p



. (17)

5.1.1 Deﬁnition

Deﬁnition 11 The psi or digamma function denoted Ψ(x) is deﬁned for any non nul or negative integer by the logarithmic derivative of Γ(x), that is :

Ψ(x) = d

dx(log(Γ(x))) . By diﬀerentiating the series (17) we ﬁnd

Ψ(x) =Γ(x)

Γ(x) =−γ − 1 x+

p=1

1 p− 1

x + p

 ,

=−γ +

p=1

1

p− 1

x + p − 1



x = 0, −1, −2, ... (18)

=−γ + p=1

 x − 1 p(x + p − 1)



x = 0, −1, −2, ...

and those series are slowly converging for any non negative integer x.

5.1.2 Properties

Polygamma functions Now if we go on diﬀerentiating relation (18) several times, we ﬁnd

Ψ(x) = Γ(x)Γ(x) − Γ2(x)

Γ2(x) =

p=1

1

(p + x − 1)2, (19) Ψ(x) = −

p=1

2 (p + x − 1)3, Ψ(n)(x) =

p=1

(−1)n+1n!

(p + x − 1)n+1, (20)

and the Ψn= Ψ(n)functions are the polygamma functions : Ψn(x) = dn+1

dxn+1(log(Γ(x))) , Ψ0(x) = Ψ(x).

Observe from (19) that for x > 0, Ψ(x) > 0 so it’s a monotonous function on the positive axis and therefore the function log(Γ(x)) is convex when x > 0.

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Recurrence relations The structure of the series expansion (18) suggests to study

Ψ(x + 1) − Ψ(x) = p=1

 1

x + p − 1− 1 x + p



which gives, just like for the gamma function, the recurrence formulae Ψ(x + 1) = Ψ(x) + 1

x, Ψ(x + n) = Ψ(x) +1

x+ 1

x + 1+ ... + 1

x + n − 1 n ≥ 1, and by diﬀerentiating the ﬁrst of those relations we deduce

Ψn(x + 1) = Ψn(x) +(−1)nn!

xn+1 . (21)

Complement and duplication formulae By logarithmic diﬀerentiation of the corresponding complement (11) and duplication (12) formulae for the gamma function we ﬁnd directly :

Theorem 12

Ψ(1− x) = Ψ(x) + π cot πx, Ψ(2x) = 1

2Ψ(x) +1 2Ψ

 x + 1

2



+ log(2).

5.1.3 Special values of the Ψn

Values at integer arguments From the relations (18) and (20) comes Ψ(1) =−γ,

Ψ1(1) = ζ(2) = π2/6, Ψ2(1) =−2ζ(3),

Ψn(1) = (−1)n+1n!ζ(n + 1), (22) where ζ(k) is the classical Riemann zeta function. Using the recurrence rela- tions (21) allow to compute those values for any other positive integer and, for example, we have

Ψ(n) = Γ(n)

Γ(n) =−γ +n−1

p=1

1

p (23)

=−γ + Hn−1.

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Values at rational arguments The value Ψ(1/2) can be computed directly from (18) or from the psi duplication formula with x = 1/2 :

Ψ

1 2



=−γ − 2 + 2 p=1

1 2p− 1

2p + 1

 ,

=−γ − 2 + 2 (1 − log(2)) = −γ − 2 log(2).

To end this section we give the interesting identities

Ψ

1 3



=−γ −3

2log(3)

3 6 π, Ψ

1 4



=−γ − 3 log(2) −π 2, Ψ

1 6



=−γ − 2 log(2) −3

2log(3)

3 2 π.

which are consequences of a more general and remarkable result : Theorem 13 (Gauss) Let 0 < p < q being integers

Ψ

p q



=−γ −π 2 cot

πp q



− log (2q) +

q−1

k=1

cos

2πkp q

 log

 sin

πk q



.

Proof. See [2] for a proof.

From this aesthetic relation, we see that the computation of Ψ(p/q) for any rational argument always involves the three fundamental mathematical con- stants : π, γ, log(2) !

5.1.4 Series expansions of the digamma function

The following series expansions are easy consequences of relations (22) and of the series

1

1 + x− 1 = −

k=2

(−1)kxk−1.

Theorem 14 (Digamma series)

Ψ(1 + x) = −γ + k=2

(−1)kζ(k)xk−1 |x| < 1, (24)

Ψ(1 + x) = − 1

1 + x− (γ − 1) +

k=2

(−1)k(ζ(k) − 1) xk−1 |x| < 1. (25)

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5.1.5 Zeros of the digamma function

The zeros of the digamma function are the extrema of the gamma function.

From the two relations

Ψ(1) =−γ < 0 Ψ(2) = 1− γ > 0,

and because Ψ(x) > 0, we see that the only positive zero x0 of the digamma function is in ]1, 2[ and its ﬁrst 50 digits are :

x0= 1.46163214496836234126265954232572132846819620400644...

Γ(x0) = 0.88560319441088870027881590058258873320795153366990..., it was ﬁrst computed by Gauss, Legendre [11] and given in [13]. On the negative axis, the digamma function has a single zero between each consecutive negative integers (the poles of the gamma function), the ﬁrst one up to 50 decimal places are





x1=−0.504083008264455409258269304533302498955385182368579...

x2=−1.573498473162390458778286043690434612655040859116846...

x3=−2.610720868444144650001537715718724207951074010873480...

x4=−3.635293366436901097839181566946017713948423861193530...

x5=−4.653237761743142441714598151148207363719069416133868...

and Hermite (1881) observed that when n becomes large [1]

xn=−n + 1 log(n)+ o

 1

log2(n)

 .

–4 –3 –2 –1 0 1 2 3 4

–4 –3 –2 –1 1 2 3 4

x

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Ψ(x) function (digamma)

### 5.2Series expansion of the gamma function

Finding series expansions for the gamma function is now an easy consequence of the series expansions for the digamma function.

Theorem 15

log(Γ(1 + x)) = −γx + k=2

(−1)kζ(k)

k xk, |x| < 1, (26)

log(Γ(1 + x)) = − log(1 + x) − (γ − 1)x + k=2

(−1)k(ζ(k) − 1)

k xk, |x| < 1.

(27) Proof. Use the term by term integration of the Taylor series (24) and (25).

We may observe that the Riemann zeta function at integer values appears in the series expansion of the logarithm of the gamma function. The convergence of the series can be accelerated by computing

1

2(log(Γ(1 + x)) − log(Γ(1 − x))) = −1 2log

1 + x 1− x



−(γ−1)x−

k=2

(ζ(2k + 1) − 1) 2k + 1 x2k+1, We now observe that the complement formula (11) becomes

Γ(1 + x)Γ(1 − x) = πx sin πx and by taking the logarithms ﬁnally

1

2log (Γ(1 + x)) +1

2(log Γ(1− x)) = 1 2log

πx

sin πx

and therefore we obtain the fast converging series due to Legendre :

log (Γ(1 + x)) = 1 2log

πx

sin πx 1

2log

1 + x 1− x



−(γ−1)x−

k=1

(ζ(2k + 1) − 1) 2k + 1 x2k+1,

(28) valid for|x| < 1.

Gauss urged to his calculating prodigy student Nicolai (1793-1846) to com- pute tables of log (Γ(x)) with twenty decimal places [8]. More modern tables related to Γ(x) and Ψ(x) are available in [1].

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### 6Euler’s constant and the gamma function

For x = 1 the formula (23) for Ψ(n) yields

Ψ(1) = Γ(1) =−γ − 1 + H1=−γ,

so Euler’s constant is the opposite of the derivative of the gamma function at x = 1.

### 6.1Euler-Mascheroni Integrals

Using the integral representation of Γ(x) gives the interesting integral formula for Euler’s constant

Γ(1) =



0 e−tlog(t)dt = −γ and from (19) comes

Ψ(1)Γ2(1) + Γ2(1) = Γ(1)Γ(1) hence

Γ(1) =



0 e−tlog2(t)dt = γ2+π2 6 .

We may go on like this and compute the Euler-Mascheroni integrals Γ(3)(1) =−γ31

2π2γ − 2ζ(3), Γ(4)(1) = γ4+ π2γ2+ 8ζ(3)γ + 3

20π4, Γ(5)(1) =−γ55

3π2γ3− 20ζ(3)γ23

4π4γ − 24ζ(5) − 10 3 ζ(3)π2, ...

### 6.2Euler’s constant and the zeta function at integer val-ues

Series formulas involving ζ(k) can also be deduced from formula (26). Taking x = 1 gives

log(Γ(2)) =−γ +

k=2

(−1)kζ(k)

k ,

thus

γ = k=2

(−1)kζ(k)

k ,

which is due to Euler. Setting x = 1/2 into (26) gives

log

 Γ

3 2



= log(

π/2) = −γ 2 +

k=2

(−1)kζ(k) k

1 2k,

(16)

therefore

γ = log

4 π

 + 2

k=2

(−1)kζ(k) 2kk. It is of interest to use the series expansion (28) at x = 1/2, log (Γ(3/2)) = 1

2log π

2 1

2log (3)1

2(γ − 1) − k=1

(ζ(2k + 1) − 1) 2k + 1

1 22k+1. It follows a fast converging expansion for γ

γ = 1 − log

3 2



k=1

(ζ(2k + 1) − 1) (2k + 1)4k . and for large values of k, we have

ζ(2k + 1) − 1 = 1

22k+1+ 1

32k+1+· · · ∼ 1

22k+1 hence ζ(2k + 1) − 1

4k 1

2 1 16k. This expression was used by Thomas Stieltjes (1856-1894) in 1887 to compute Euler’s constant up to 32 decimal places [14]. In the same article he also com- puted ζ(2) up to ζ(70) with 32 digits.

### 7The gamma function and the Riemann Zetafunction

The integral deﬁnition of the gamma function Γ(x) =



0 tx−1e−tdt,

together with the change of variables t = ku (with k a positive integer) yields Γ(x) =



0 (ku)x−1e−kuk du = kx



0 ux−1e−kudu.

We write this in the form 1 kx = 1

Γ(x)



0 ux−1e−kudu, hence by summation

k=1

1 kx = 1

Γ(x)



0 ux−1 k=1

e−ku du

= 1

Γ(x)



0 ux−1

 1

1− e−u − 1

 du.

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We have obtained the beautiful formula ζ(x)Γ(x) =



0

tx−1

et− 1dt (29)

and, for example, for x = 2, (29) becomes π2

6 =



0

t et− 1dt.

There is another celebrated and most important functional equation between those two functions, the Riemann zeta function functional equation :

Theorem 16 (Riemann, 1859) Let

Λ(s) = π−s/2Γ s

2

ζ(s), an analytic function except at poles 0 and 1, then

Λ(s) = Λ(1 − s).

Proof. Several proofs may be found in [15]. Euler demonstrated it for integer values of s.

This equation allows to extend the deﬁnition of the Zeta function to negative values of the arguments.

### 8The Beta function

Let us now consider the useful and related function to the gamma function which occurs in the computation of many deﬁnite integrals. It’s deﬁned, for x > 0 and y > 0 by the two equivalent identities :

Deﬁnition 17 The beta function (or Eulerian integral of the ﬁrst kind) is given by

B(x, y) =

 1

0 tx−1(1− t)y−1dt (30)

= 2

 π/2

0 sin(t)2x−1cos(t)2y−1dt = 2

 π/2

0 sin(t)2y−1cos(t)2x−1dt, (31)

= B(y, x).

This deﬁnition is also valid for complex numbers x and y such as (x) >

0 and (y) > 0 and Euler gave (30) in 1730. The name beta function was introduced for the ﬁrst time by Jacques Binet (1786-1856) in 1839 [5] and he made various contributions on the subject.

The beta function is symmetric and may be computed by mean of the gamma function thanks to the important property :

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Theorem 18 Let (x) > 0 and (y) > 0, then B(x, y) =Γ(x)Γ(y)

Γ(x + y) = B(y, x). (32)

Proof. We use the deﬁnite integral (3) and form the following product

Γ(x)Γ(y) = 4



0 u2x−1e−u2du



0 v2y−1e−v2dv

= 4



0



0 e(u2+v2)u2x−1v2y−1dudv, we introduce the polar variables u = r cos θ, v = r sin θ so that

Γ(x)Γ(y) = 4



0

 π/2

0 e−r2r2(x+y)−1cos2x−1θ sin2y−1θdrdθ

= 2



0 r2(x+y)−1e−r2dr.2

 π/2

0

cos2x−1θ sin2y−1θdθ

= Γ(x + y)B(y, x).

From relation (32) follows B(x + 1, y) = Γ(x + 1)Γ(y)

Γ(x + y + 1) = xΓ(x)Γ(y)

(x + y)Γ(x + y) = x

x + yB(x, y), this is the beta function functional equation

B(x + 1, y) = x

x + yB(x, y). (33)

### 8.1Special values

B

1 2,1

2



= π,

B

1 3,2

3



=2 3 3 π, B

1 4,3

4



= π√ 2, B(x, 1 − x) = π

sin(πx), B(x, 1) = 1

x,

B(x, n) = (n − 1)!

x.(x + 1)...(x + n − 1) n ≥ 1, B(m, n) = (m − 1)!(n − 1)!

(m + n − 1)! m ≥ 1, n ≥ 1.

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### 8.2Wallis’s integrals

For example the following integrals (Wallis’s integrals) Wn =

 π/2

0

sinnθdθ =

 π/2

0

cosnθdθ,

may be computed by mean of the beta and gamma functions. Thanks to the relation (31), we have

Wn =1 2B

n + 1 2 ,1

2

 ,

and come naturally the two cases n = 2p + 1 and n = 2p. For the odd values of the argument n :

W2p+1= 1 2B

 p + 1,1

2



=Γ(p + 1)Γ(1/2)

2Γ(p + 3/2) = p!Γ(1/2) (2p + 1)Γ(p + 1/2) and using formula (10) produces the well-known result

W2p+1= 2pp!

1.3.5...(2p + 1)= 4pp!2 (2p + 1)!.

The same method permits to compute the integrals for the even values W2p= 1

2B

 p + 1

2,1 2



= Γ(p + 1/2)Γ(1/2) 2Γ(p + 1) and ﬁnally

W2p =1.3.5...(2p − 1)

2p+1p! π = (2p)!

4pp!2 π 2. Observe that it’s easy to see that

Wn+2= 1 2B

n + 2 + 1 2 ,1

2



= 1 2B

n + 1 2 + 1,1

2



=(n + 1)/2 n/2 + 1 Wn =

n + 1 n + 2

 Wn

thanks to the beta function functional equation (33).

It’s interesting to notice that Wα=1

2B

α + 1 2 ,1

2



also works for any real number α > −1 and therefore we may deduce using (30) and (31) that (respectively with α = −1/2 and α = 1/2)

 π/2

0

√dθ sin θ =

 1 0

√2dt

1− t4 = Γ2(1/4) 2

, (34)

 π/2

0

√sin θdθ =

 1 0

2t2dt

1− t4 = (2π)3/2 Γ2(1/4).

Consequently the product of those two integrals permits to derive the relation

due to Euler  1

0

dt 1− t4

 1 0

t2dt

1− t4 = π 4.

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### References

[1] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions, Dover, New York, (1964)

[2] G.E. Andrews, R. Askey and R. Roy, Special functions, Cambridge Univer- sity Press, Cambridge, (1999)

[3] E. Artin, The Gamma Function, New York, Holt, Rinehart and Winston, (1964)

[4] E.W. Barnes, The theory of the gamma function, Messenger Math. (2), (1900), vol. 29, p. 64-128.

[5] J.P.M Binet, Journal ´ecole polyt., (1839), vol. 16, p. 131

[6] H. Bohr and I. Mollerup, Loerbog I matematisk Analyse, Kopenhagen, (1922), vol. 3

[7] J.M. Borwein and I.J. Zucker, Elliptic integral evaluation of the Gamma function at rational values of small denominator, IMA J. of Numer Anal- ysis, (1992), vol. 12, p .519-526

[8] C.F. Gauss, Werke, G¨ottingen, (1866-1933), vol. 3

[9] M. Godefroy, La fonction Gamma ; Th´eorie, Histoire, Bibliographie, Gauthier-Villars, Paris, (1901)

[10] X. Gourdon and P. Sebah, Numbers, Constants and Computation, World Wide Web site at the adress : http://numbers.computation.free.fr/Constants/constants.html, (1999)

[11] A.M. Legendre, M´emoires de la classe des sciences math´ematiques et physiques de l’Institut de France, Paris, (1809), p. 477, 485, 490

[12] N. Nielsen, Handbuch der Theorie der Gammafunktion, Leipzig, (1906) [13] W. Sibagaki, Theory and applications of the gamma function, Iwanami

Syoten, Tokyo, Japan, (1952)

[14] T.J. Stieltjes, Tables des valeurs des sommes Sk =

n=1n−k,Acta Math- ematica, (1887), vol. 10, p. 299-302

[15] E.C. Titchmarsh, The theory of the Riemann Zeta-function, Oxford Science publications, second edition, revised by D.R. Heath-Brown (1986)

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