the gamma function - david altizio's web page

The Gamma Function

David Altizio

CMU Math Club

31 January 2020

Introduction

“[The Gamma function is] arguably, the most common special function, or the

least ‘special’ of them. The other transcendental functions...are called ‘special’

because you could conceivably avoid some of them by staying away from many

specialized mathematical topics. On the other hand, the Gamma function

y = Γ(x) is most difficult to avoid.” -Gerard P. Michon

Probability: the gamma distribution can be used to modeltime-based occurrences, such as the life length of an electroniccomponent

Physics: useful tool for string theory, Feynman diagrams,Maxwell-Boltzmann statistics, and so on

Analysis: shows up in the Poisson kernel and the definition ofHausdorff measure (useful for analyzing fractals!)

Probably others....

So what’s all the fuss about?

Introduction

Probably others....

Introduction

Probably others....

Introduction

Probably others....

Introduction

Probably others....

Introduction

Probably others....

The Basics

Definition

For positive real numbers x , the gamma function Γ(x) is defined via

Γ(x) :=

∫ ∞0

tx−1e−t dt.

The restriction on x is important: near zero, tx−1e−t is basically tx−1,and in order for this to be integrable, we need x > 0.

This definition carries over to complex numbers z with ease; the domainrestriction then changes to <(z) > 0.

The Basics

Definition

Γ(x) :=

∫ ∞0

tx−1e−t dt.

The Basics

Definition

Γ(x) :=

∫ ∞0

tx−1e−t dt.

The Basics

Let x > 0. A quick application of Integration by Parts yields∫ ∞0

txe−t dt = x

∫ ∞0

tx−1e−t dt,

orΓ(x + 1) = xΓ(x).

Combined with Γ(1) =∫∞

0e−t dt = 1, we deduce

Γ(n) = (n − 1)! for each integer n ≥ 1.

Thus, the gamma function can be considered a “continuous extension” ofthe factorial function.

Theorem (Bohr–Mollerup, 1922)

The function y = Γ(x) is the only continuous extension of the factorialfunction such that log(Γ(x)) is convex.

The Basics

txe−t dt = x

∫ ∞0

tx−1e−t dt,

orΓ(x + 1) = xΓ(x).

The Basics

txe−t dt = x

∫ ∞0

tx−1e−t dt,

orΓ(x + 1) = xΓ(x).

The Basics

We can also compute Γ( 12 ): the u-substitution t = u2 (so that

dt = 2u du) yields

Γ( 12 ) =

∫ ∞0

t−1/2e−t dt = 2

∫ ∞0

e−u2

du =√π.

We may thus compute, for example,

Γ( 52 ) = 3

2 Γ( 32 ) = 3

2 ·12 Γ( 1

2 ) = 34

√π.

Unfortunately, this is about as good as we can get in terms of computingexplicit values of Γ(x). For example, no closed form of Γ( 1

3 ) or Γ( 23 ) is

known, although we do know that both numbers are transcendental.

The Basics

dt = 2u du) yields

Γ( 12 ) =

∫ ∞0

t−1/2e−t dt = 2

∫ ∞0

e−u2

du =√π.

Γ( 52 ) = 3

2 Γ( 32 ) = 3

2 ·12 Γ( 1

2 ) = 34

√π.

3 ) or Γ( 23 ) is

The Basics

dt = 2u du) yields

Γ( 12 ) =

∫ ∞0

t−1/2e−t dt = 2

∫ ∞0

e−u2

du =√π.

Γ( 52 ) = 3

2 Γ( 32 ) = 3

2 ·12 Γ( 1

2 ) = 34

√π.

3 ) or Γ( 23 ) is

Application 1: Volume of Unit Ball

For each integer n ≥ 1, set

Bn := {~x = (x1, . . . , xn) ∈ Rn : |x1|2 + · · ·+ |xn|2 ≤ 1}

to be the n-dimensional unit ball in Rn. Similarly, define

Sn−1 := {~x = (x1, . . . , xn) ∈ Rn : |x1|2 + · · ·+ |xn|2 = 1}

to be the n-dimensional unit sphere in Rn. We know, for instance, thatthe volume of B2 is π while the surface area of S1 is 2π. But can wecompute the volume of Bn and the surface area of Sn−1 in general?

Theorem

For each n ≥ 1, let ωn be the volume of Bn, and let αn be the surfacearea of Sn−1. Then

αn =2πn/2

Γ( n2 )

and ωn =πn/2

Γ( n2 + 1)

For each integer n ≥ 1, set

Bn := {~x = (x1, . . . , xn) ∈ Rn : |x1|2 + · · ·+ |xn|2 ≤ 1}

to be the n-dimensional unit ball in Rn. Similarly, define

Sn−1 := {~x = (x1, . . . , xn) ∈ Rn : |x1|2 + · · ·+ |xn|2 = 1}

to be the n-dimensional unit sphere in Rn. We know, for instance, thatthe volume of B2 is π while the surface area of S1 is 2π. But can wecompute the volume of Bn and the surface area of Sn−1 in general?

Theorem

For each n ≥ 1, let ωn be the volume of Bn, and let αn be the surfacearea of Sn−1. Then

αn =2πn/2

Γ( n2 )

and ωn =πn/2

Γ( n2 + 1)

Proof. Define the function f : Rn → R via

f (x1, . . . , xn) = e−|~x|2

= e−x21 e−x

22 · · · e−x

We’ll compute the integral∫Rn f dV in two different ways.

On one hand, note that f is the product of n different functions, eachdepending on a different variable. Thus, we may split up the integral(Fubini!) and write∫

f dV =

e−x21 e−x

22 · · · e−x

2n dxn · · · dx2 dx1

(∫ ∞−∞

e−x2j dxj

)= πn/2.

Proof. Define the function f : Rn → R via

f (x1, . . . , xn) = e−|~x|2

= e−x21 e−x

22 · · · e−x

We’ll compute the integral∫Rn f dV in two different ways.

On one hand, note that f is the product of n different functions, eachdepending on a different variable. Thus, we may split up the integral(Fubini!) and write∫

f dV =

e−x21 e−x

22 · · · e−x

2n dxn · · · dx2 dx1

(∫ ∞−∞

e−x2j dxj

)= πn/2.

On the other hand, we may use spherical coordinates and write∫Rn

f dV =

∫ ∞0

∫{|~x|=r}

e−|~x|2

∫ ∞0

(rn−1αn)e−r2

∫ ∞0

n2−1e−u du = αn · 1

2 Γ(n2

Solving yields αn = 2πn/2

Γ( n2 ) .

We may then compute

rn−1αn dr =2πn/2

nΓ( n2 )

=πn/2

Γ( n2 + 1)

f dV =

∫ ∞0

∫{|~x|=r}

e−|~x|2

∫ ∞0

(rn−1αn)e−r2

∫ ∞0

n2−1e−u du = αn · 1

2 Γ(n2

Γ( n2 ) .

We may then compute

nΓ( n2 )

=πn/2

Γ( n2 + 1)

f dV =

∫ ∞0

∫{|~x|=r}

e−|~x|2

∫ ∞0

(rn−1αn)e−r2

∫ ∞0

n2−1e−u du = αn · 1

2 Γ(n2

Γ( n2 ) .

We may then compute

nΓ( n2 )

=πn/2

Γ( n2 + 1)

f dV =

∫ ∞0

∫{|~x|=r}

e−|~x|2

∫ ∞0

(rn−1αn)e−r2

∫ ∞0

n2−1e−u du = αn · 1

2 Γ(n2

Γ( n2 ) .

We may then compute

nΓ( n2 )

=πn/2

Γ( n2 + 1)

f dV =

∫ ∞0

∫{|~x|=r}

e−|~x|2

∫ ∞0

(rn−1αn)e−r2

∫ ∞0

n2−1e−u du = αn · 1

2 Γ(n2

Γ( n2 ) .

We may then compute

nΓ( n2 )

=πn/2

Γ( n2 + 1)

More generally, for positive real numbers p1, . . . , pn, the volume of then-dimensional region

Bp1,...,pn := {~x = (x1, . . . , xn) ∈ Rn : |x1|p1 + · · ·+ |xn|pn ≤ 1}

vol(Bp1,...,pn) = 2nΓ(1 + 1

p1) · · · Γ(1 + 1

Γ(1 + 1p1

+ · · ·+ 1pn

Product Construction

It turns out that some “nice” functions can be written as convergentinfinite products. For example, one can show that

sin(πz) = πz∞∏n=1

(1− z2

)for every complex number z .

The Gamma function is one such function: indeed, with some work, it ispossible to prove that

Γ(z)= e−γzz

∞∏n=1

)e−z/n,

where γ is the Euler-Mascheroni constant. (The e−z/n term is needed tomake the product converge nicely.)

This also gives us a way to define Γ(z) for all complex numbersz ∈ C \ {0,−1,−2, . . .}.

(1− z2

Γ(z)= e−γzz

∞∏n=1

)e−z/n,

(1− z2

Γ(z)= e−γzz

∞∏n=1

)e−z/n,

Reflection Formula

This formulation of Γ(z) has some distinct advantages. For example,observe that

Γ(z)Γ(−z)=

(e−γzz

∞∏n=1

)e−z/n

)(−eγzz

∞∏n=1

(1− z

= −z2∞∏n=1

(1− z2

)=−z sin(πz)

From the relation Γ(1− z) = −zΓ(−z), we thus deduce the Reflectionformula

Γ(z)Γ(1− z) =π

sin(πz).

This is very difficult to deduce using the integral definition of Γ!

Reflection Formula

This formulation of Γ(z) has some distinct advantages. For example,observe that

Γ(z)Γ(−z)=

(e−γzz

∞∏n=1

)e−z/n

)(−eγzz

∞∏n=1

(1− z

= −z2∞∏n=1

(1− z2

)=−z sin(πz)

From the relation Γ(1− z) = −zΓ(−z), we thus deduce the Reflectionformula

Γ(z)Γ(1− z) =π

sin(πz).

This is very difficult to deduce using the integral definition of Γ!

Application 2: The Zeta Function

For complex numbers s with <(s) > 1, define the zeta function ζ(s) via

ζ(s) :=1

2s+ · · · =

∞∑k=1

An application of unique prime factorization allows us to write

ζ(s) =∏

p prime

p2s+ · · ·

∏p prime

1− p−s;

this makes the zeta function crucial in the study of modern numbertheory.

Some common values: ζ(2) = π2

6 , ζ(4) = π4

90 , ζ(3) =???.

ζ(s) :=1

2s+ · · · =

∞∑k=1

ζ(s) =∏

p prime

p2s+ · · ·

∏p prime

1− p−s;

6 , ζ(4) = π4

90 , ζ(3) =???.

ζ(s) :=1

2s+ · · · =

∞∑k=1

ζ(s) =∏

p prime

p2s+ · · ·

∏p prime

1− p−s;

6 , ζ(4) = π4

90 , ζ(3) =???.

Let s be a complex number with <(s) > 1. For each positive integer n,the u-substitution t = nu implies

Γ(s) =

∫ ∞0

ts−1e−t dt = ns∫ ∞

us−1e−nu du.

Dividing both sides by ns and summing over all n yields

ζ(s)Γ(s) =∞∑n=1

n−sΓ(s) =∞∑n=1

∫ ∞0

us−1e−nu du

∫ ∞0

∞∑n=1

us−1e−nu du =

∫ ∞0

us−1

eu − 1du.

Hence the zeta and gamma functions are connected in an integral way(pun intended)!This means it may be possible to (analytically!) define ζ(s) for othercomplex numbers s.

Γ(s) =

∫ ∞0

us−1e−nu du.

ζ(s)Γ(s) =∞∑n=1

n−sΓ(s) =∞∑n=1

∫ ∞0

us−1e−nu du

∫ ∞0

∞∑n=1

us−1e−nu du =

∫ ∞0

us−1

eu − 1du.

Hence the zeta and gamma functions are connected in an integral way(pun intended)!

This means it may be possible to (analytically!) define ζ(s) for othercomplex numbers s.

Γ(s) =

∫ ∞0

us−1e−nu du.

ζ(s)Γ(s) =∞∑n=1

n−sΓ(s) =∞∑n=1

∫ ∞0

us−1e−nu du

∫ ∞0

∞∑n=1

us−1e−nu du =

∫ ∞0

us−1

eu − 1du.

Hence the zeta and gamma functions are connected in an integral way(pun intended)!This means it may be possible to (analytically!) define ζ(s) for othercomplex numbers s.

With a lot of work, we may derive the Riemann functional equation

ζ(s) = 2sπs−1 sin(πs

)Γ(1− s)ζ(1− s),

which holds for every s. (The product Γ(1− s)ζ(1− s) on the right handside comes from the equality on the previous page.)

For example, plugging in s = −1 yields

ζ(−1) = 2−1π−2 sin(−π

)Γ(2)ζ(2) = − 1

This is what the “equality”

1 + 2 + 3 + · · · = − 112

actually means.

With a lot of work, we may derive the Riemann functional equation

ζ(s) = 2sπs−1 sin(πs

)Γ(1− s)ζ(1− s),

which holds for every s. (The product Γ(1− s)ζ(1− s) on the right handside comes from the equality on the previous page.)

For example, plugging in s = −1 yields

ζ(−1) = 2−1π−2 sin(−π

)Γ(2)ζ(2) = − 1

This is what the “equality”

1 + 2 + 3 + · · · = − 112

actually means.

Application 3: Integration

One of the most important properties of the gamma function is thefollowing result.

Theorem (Beta Integral)

Let p > 0 and q > 0 be real numbers. Then∫ 1

tp−1(1− t)q−1 dt =Γ(p)Γ(q)

Γ(p + q).

Proof. Write

Γ(p)Γ(q) =

(∫ ∞0

xp−1e−x dx

)(∫ ∞0

yq−1e−y dy

∫ ∞0

xp−1yq−1e−(x+y) dx dy .

One of the most important properties of the gamma function is thefollowing result.

Theorem (Beta Integral)

Let p > 0 and q > 0 be real numbers. Then∫ 1

tp−1(1− t)q−1 dt =Γ(p)Γ(q)

Γ(p + q).

Proof. Write

Γ(p)Γ(q) =

(∫ ∞0

xp−1e−x dx

)(∫ ∞0

yq−1e−y dy

∫ ∞0

xp−1yq−1e−(x+y) dx dy .

Now make the change of variables x = tu, y = (1− t)u, for t ∈ (0, 1)and u ∈ (0,∞). The Jacobian of this linear transformation is u, and sothe integral equals∫ ∞

(tu)p−1((1− t)u)q−1e−t · u dt du

∫ ∞0

up+q−1tp−1(1− t)q−1e−u dt du

(∫ ∞0

up+q−1e−u du

)(∫ 1

tp−1(1− t)q−1 dt

)= Γ(p + q)

tp−1(1− t)q−1 dt.

Dividing both sides by Γ(p + q) 6= 0 yields the desired.

∫ ∞0

(∫ ∞0

up+q−1e−u du

)(∫ 1

tp−1(1− t)q−1 dt

)= Γ(p + q)

tp−1(1− t)q−1 dt.

∫ ∞0

(∫ ∞0

up+q−1e−u du

)(∫ 1

tp−1(1− t)q−1 dt

= Γ(p + q)

tp−1(1− t)q−1 dt.

∫ ∞0

(∫ ∞0

up+q−1e−u du

)(∫ 1

tp−1(1− t)q−1 dt

)= Γ(p + q)

tp−1(1− t)q−1 dt.

∫ ∞0

(∫ ∞0

up+q−1e−u du

)(∫ 1

tp−1(1− t)q−1 dt

)= Γ(p + q)

tp−1(1− t)q−1 dt.

We can use this to compute ∫ 1

dx√1− x1/4

Indeed, making the u-substitution u = x1/4 (so that dx = 4u3 du) yields∫ 1

dx√1− x1/4

4u3(1− u)−1/2 du =4Γ(4)Γ( 1

Γ( 92 )

But we already know that Γ(4) = 3! = 6 and

Γ( 92 ) = 7

2 ·52 ·

12 Γ( 1

2 ) = 10516 Γ( 1

Therefore the above expression simplifies nicely to

4 · 6 · Γ( 12 )

10516 Γ( 1

dx√1− x1/4

4u3(1− u)−1/2 du =4Γ(4)Γ( 1

Γ( 92 )

Γ( 92 ) = 7

2 ·52 ·

12 Γ( 1

2 ) = 10516 Γ( 1

4 · 6 · Γ( 12 )

10516 Γ( 1

dx√1− x1/4

4u3(1− u)−1/2 du =4Γ(4)Γ( 1

Γ( 92 )

Γ( 92 ) = 7

2 ·52 ·

12 Γ( 1

2 ) = 10516 Γ( 1

4 · 6 · Γ( 12 )

10516 Γ( 1

Let’s do a wackier example and compute∫ 1

dx3√x2(1− x)

In the same vein as before, this rewrites as∫ 1

dx3√x2(1− x)

x−2/3(1− x)−1/3 dx =Γ( 1

3 )Γ( 23 )

Γ(1).

Note that Γ(1) = 1. Moreover, although Γ( 13 ) and Γ( 2

3 ) are hopelesslycomplicated, the Reflection Formula

Γ(x)Γ(1− x) =π

sin(πx)

tells us their product is πsin(π3 ) = 2π√

Hence the desired integral equals 2π√3

dx3√x2(1− x)

x−2/3(1− x)−1/3 dx =Γ( 1

3 )Γ( 23 )

Γ(1).

Γ(x)Γ(1− x) =π

sin(πx)

dx3√x2(1− x)

x−2/3(1− x)−1/3 dx =Γ( 1

3 )Γ( 23 )

Γ(1).

Γ(x)Γ(1− x) =π

sin(πx)

dx3√x2(1− x)

x−2/3(1− x)−1/3 dx =Γ( 1

3 )Γ( 23 )

Γ(1).

Γ(x)Γ(1− x) =π

sin(πx)

Application 4: Sums

Recall that the binomial coefficient(nk

)is computed via(

(n − k)!k!.

Of interest to us are the central binomial coefficients(

In combinatorics, power series are very useful tools for either solving orestimating the terms of recursions. One useful power series is

∞∑n=0

1√1− 4z

this follows by writing the right hand side as (1− 4z)−1/2 and using thegeneralized binomial theorem.

But what about the sum∞∑n=0

zn(2nn

Application 4: Sums

)is computed via(

(n − k)!k!.

∞∑n=0

1√1− 4z

zn(2nn

Application 4: Sums

)is computed via(

(n − k)!k!.

∞∑n=0

1√1− 4z

zn(2nn

Application 4: Sums

To motivate the following discussion, write

(2n)!n!n!

(2n)!.

The form of this fraction highly suggests we can write it as a Betaintegral. However, this doesn’t quite work, as

(2n)!=

Γ(n + 1)Γ(n + 1)

Γ(2n + 1),

which doesn’t quite fit the mold we’re looking for.

That being said, we can extract a Beta integral if the fraction we’re

interested in is n!(n−1)!(2n)! . This can be achieved by integrating our power

series!

Application 4: Sums

(2n)!n!n!

(2n)!.

(2n)!=

Γ(n + 1)Γ(n + 1)

Γ(2n + 1),

series!

Application 4: Sums

(2n)!n!n!

(2n)!.

(2n)!=

Γ(n + 1)Γ(n + 1)

Γ(2n + 1),

series!

Application 4: Sums

Define the function F (z) via

F (z) :=∞∑n=1

n!(n − 1)!

(2n)!zn.

A simple application of the ratio test tells us that the sum on the rightconverges for all |z | < 4; this means F ′(z) is defined for |z | < 4.

(n − 1)!n!

(2n)!zn =

Γ(n)Γ(n + 1)

Γ(2n + 1)zn

tn−1((1− t)z)n dt

for every integer n ≥ 1.

Application 4: Sums

Define the function F (z) via

F (z) :=∞∑n=1

n!(n − 1)!

(2n)!zn.

A simple application of the ratio test tells us that the sum on the rightconverges for all |z | < 4; this means F ′(z) is defined for |z | < 4.

(n − 1)!n!

(2n)!zn =

Γ(n)Γ(n + 1)

Γ(2n + 1)zn

tn−1((1− t)z)n dt

for every integer n ≥ 1.

Application 4: Sums

Absolute convergence allows us to swap the order of the summation andthe integral, and so within a neighborhood of z = 1 we have

∞∑n=1

n!(n − 1)!

(2n)!zn =

∞∑n=1

tn−1(z(1− t))n dt

∞∑n=1

tn−1(z(1− t))n dt

z(1− t)

1− tz(1− t)dt =

1− t

t2 − t + 1z

A bit of elbow grease (try it; it’s not as scary as it looks!) shows that this

integral equals√

z4−z arcsin(

Application 4: Sums

Differentiating both sides with respect to z yields

∞∑n=1

(2n)!zn = F ′(z) =

4− z+

(4− z)3/2arcsin

Hence the value of the sum is 1 + F ′(z). In particular,

∞∑n=0

) = 1 + F ′(1) = 1 +1

33/2· π

36 + 2π√

Application 4: Sums

Differentiating both sides with respect to z yields

∞∑n=1

(2n)!zn = F ′(z) =

4− z+

(4− z)3/2arcsin

Hence the value of the sum is 1 + F ′(z). In particular,

∞∑n=0

) = 1 + F ′(1) = 1 +1

33/2· π

36 + 2π√

Some Extra Problems

Problem 1 (Classic?)

Let Cn := 1n+1

)be the nth Catalan number. Prove that

∞∑n=0

Cn= 2 +

Problem 2 (AMM 11103, Gregory Galperin and Hillel Gauchman)

Prove that for every positive integer n,

n∑k=1

2n−1

n∑k=1k odd

the gamma function - david altizio's web page

Documents