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
“[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
“[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
“[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
“[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
“[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?
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
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
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
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
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
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
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
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
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.
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)
.
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)
.
Application 1: Volume of Unit Ball
Proof. Define the function f : Rn → R via
f (x1, . . . , xn) = e−|~x|2
= e−x21 e−x
22 · · · e−x
2n .
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∫
Rn
f dV =
∫Rn
e−x21 e−x
22 · · · e−x
2n dxn · · · dx2 dx1
=n∏
j=1
(∫ ∞−∞
e−x2j dxj
)= πn/2.
Application 1: Volume of Unit Ball
Proof. Define the function f : Rn → R via
f (x1, . . . , xn) = e−|~x|2
= e−x21 e−x
22 · · · e−x
2n .
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∫
Rn
f dV =
∫Rn
e−x21 e−x
22 · · · e−x
2n dxn · · · dx2 dx1
=n∏
j=1
(∫ ∞−∞
e−x2j dxj
)= πn/2.
Application 1: Volume of Unit Ball
On the other hand, we may use spherical coordinates and write∫Rn
f dV =
∫ ∞0
∫{|~x|=r}
e−|~x|2
dA dr
=
∫ ∞0
(rn−1αn)e−r2
dr
= αn
∫ ∞0
12u
n2−1e−u du = αn · 1
2 Γ(n2
).
Solving yields αn = 2πn/2
Γ( n2 ) .
We may then compute
ωn =
∫ 1
0
rn−1αn dr =2πn/2
nΓ( n2 )
=πn/2
Γ( n2 + 1)
. �
Application 1: Volume of Unit Ball
On the other hand, we may use spherical coordinates and write∫Rn
f dV =
∫ ∞0
∫{|~x|=r}
e−|~x|2
dA dr
=
∫ ∞0
(rn−1αn)e−r2
dr
= αn
∫ ∞0
12u
n2−1e−u du = αn · 1
2 Γ(n2
).
Solving yields αn = 2πn/2
Γ( n2 ) .
We may then compute
ωn =
∫ 1
0
rn−1αn dr =2πn/2
nΓ( n2 )
=πn/2
Γ( n2 + 1)
. �
Application 1: Volume of Unit Ball
On the other hand, we may use spherical coordinates and write∫Rn
f dV =
∫ ∞0
∫{|~x|=r}
e−|~x|2
dA dr
=
∫ ∞0
(rn−1αn)e−r2
dr
= αn
∫ ∞0
12u
n2−1e−u du = αn · 1
2 Γ(n2
).
Solving yields αn = 2πn/2
Γ( n2 ) .
We may then compute
ωn =
∫ 1
0
rn−1αn dr =2πn/2
nΓ( n2 )
=πn/2
Γ( n2 + 1)
. �
Application 1: Volume of Unit Ball
On the other hand, we may use spherical coordinates and write∫Rn
f dV =
∫ ∞0
∫{|~x|=r}
e−|~x|2
dA dr
=
∫ ∞0
(rn−1αn)e−r2
dr
= αn
∫ ∞0
12u
n2−1e−u du = αn · 1
2 Γ(n2
).
Solving yields αn = 2πn/2
Γ( n2 ) .
We may then compute
ωn =
∫ 1
0
rn−1αn dr =2πn/2
nΓ( n2 )
=πn/2
Γ( n2 + 1)
. �
Application 1: Volume of Unit Ball
On the other hand, we may use spherical coordinates and write∫Rn
f dV =
∫ ∞0
∫{|~x|=r}
e−|~x|2
dA dr
=
∫ ∞0
(rn−1αn)e−r2
dr
= αn
∫ ∞0
12u
n2−1e−u du = αn · 1
2 Γ(n2
).
Solving yields αn = 2πn/2
Γ( n2 ) .
We may then compute
ωn =
∫ 1
0
rn−1αn dr =2πn/2
nΓ( n2 )
=πn/2
Γ( n2 + 1)
. �
Application 1: Volume of Unit Ball
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}
is
vol(Bp1,...,pn) = 2nΓ(1 + 1
p1) · · · Γ(1 + 1
pn)
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
n2
)for every complex number z .
The Gamma function is one such function: indeed, with some work, it ispossible to prove that
1
Γ(z)= e−γzz
∞∏n=1
(1 +
z
n
)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, . . .}.
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
n2
)for every complex number z .
The Gamma function is one such function: indeed, with some work, it ispossible to prove that
1
Γ(z)= e−γzz
∞∏n=1
(1 +
z
n
)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, . . .}.
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
n2
)for every complex number z .
The Gamma function is one such function: indeed, with some work, it ispossible to prove that
1
Γ(z)= e−γzz
∞∏n=1
(1 +
z
n
)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, . . .}.
Reflection Formula
This formulation of Γ(z) has some distinct advantages. For example,observe that
1
Γ(z)Γ(−z)=
(e−γzz
∞∏n=1
(1 +
z
n
)e−z/n
)(−eγzz
∞∏n=1
(1− z
n
)ez/n
)
= −z2∞∏n=1
(1− z2
n2
)=−z sin(πz)
π.
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
1
Γ(z)Γ(−z)=
(e−γzz
∞∏n=1
(1 +
z
n
)e−z/n
)(−eγzz
∞∏n=1
(1− z
n
)ez/n
)
= −z2∞∏n=1
(1− z2
n2
)=−z sin(πz)
π.
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
1s+
1
2s+ · · · =
∞∑k=1
1
ks.
An application of unique prime factorization allows us to write
ζ(s) =∏
p prime
(1 +
1
ps+
1
p2s+ · · ·
)=
∏p prime
1
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) =???.
Application 2: The Zeta Function
For complex numbers s with <(s) > 1, define the zeta function ζ(s) via
ζ(s) :=1
1s+
1
2s+ · · · =
∞∑k=1
1
ks.
An application of unique prime factorization allows us to write
ζ(s) =∏
p prime
(1 +
1
ps+
1
p2s+ · · ·
)=
∏p prime
1
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) =???.
Application 2: The Zeta Function
For complex numbers s with <(s) > 1, define the zeta function ζ(s) via
ζ(s) :=1
1s+
1
2s+ · · · =
∞∑k=1
1
ks.
An application of unique prime factorization allows us to write
ζ(s) =∏
p prime
(1 +
1
ps+
1
p2s+ · · ·
)=
∏p prime
1
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) =???.
Application 2: The Zeta Function
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∫ ∞
0
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.
Application 2: The Zeta Function
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∫ ∞
0
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.
Application 2: The Zeta Function
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∫ ∞
0
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.
Application 2: The Zeta Function
With a lot of work, we may derive the Riemann functional equation
ζ(s) = 2sπs−1 sin(πs
2
)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
2
)Γ(2)ζ(2) = − 1
12 .
This is what the “equality”
1 + 2 + 3 + · · · = − 112
actually means.
Application 2: The Zeta Function
With a lot of work, we may derive the Riemann functional equation
ζ(s) = 2sπs−1 sin(πs
2
)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
2
)Γ(2)ζ(2) = − 1
12 .
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
0
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
∫ ∞0
xp−1yq−1e−(x+y) dx dy .
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
0
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
∫ ∞0
xp−1yq−1e−(x+y) dx dy .
Application 3: Integration
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∫ ∞
0
∫ 1
0
(tu)p−1((1− t)u)q−1e−t · u dt du
=
∫ ∞0
∫ 1
0
up+q−1tp−1(1− t)q−1e−u dt du
=
(∫ ∞0
up+q−1e−u du
)(∫ 1
0
tp−1(1− t)q−1 dt
)= Γ(p + q)
∫ 1
0
tp−1(1− t)q−1 dt.
Dividing both sides by Γ(p + q) 6= 0 yields the desired.
Application 3: Integration
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∫ ∞
0
∫ 1
0
(tu)p−1((1− t)u)q−1e−t · u dt du
=
∫ ∞0
∫ 1
0
up+q−1tp−1(1− t)q−1e−u dt du
=
(∫ ∞0
up+q−1e−u du
)(∫ 1
0
tp−1(1− t)q−1 dt
)= Γ(p + q)
∫ 1
0
tp−1(1− t)q−1 dt.
Dividing both sides by Γ(p + q) 6= 0 yields the desired.
Application 3: Integration
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∫ ∞
0
∫ 1
0
(tu)p−1((1− t)u)q−1e−t · u dt du
=
∫ ∞0
∫ 1
0
up+q−1tp−1(1− t)q−1e−u dt du
=
(∫ ∞0
up+q−1e−u du
)(∫ 1
0
tp−1(1− t)q−1 dt
)
= Γ(p + q)
∫ 1
0
tp−1(1− t)q−1 dt.
Dividing both sides by Γ(p + q) 6= 0 yields the desired.
Application 3: Integration
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∫ ∞
0
∫ 1
0
(tu)p−1((1− t)u)q−1e−t · u dt du
=
∫ ∞0
∫ 1
0
up+q−1tp−1(1− t)q−1e−u dt du
=
(∫ ∞0
up+q−1e−u du
)(∫ 1
0
tp−1(1− t)q−1 dt
)= Γ(p + q)
∫ 1
0
tp−1(1− t)q−1 dt.
Dividing both sides by Γ(p + q) 6= 0 yields the desired.
Application 3: Integration
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∫ ∞
0
∫ 1
0
(tu)p−1((1− t)u)q−1e−t · u dt du
=
∫ ∞0
∫ 1
0
up+q−1tp−1(1− t)q−1e−u dt du
=
(∫ ∞0
up+q−1e−u du
)(∫ 1
0
tp−1(1− t)q−1 dt
)= Γ(p + q)
∫ 1
0
tp−1(1− t)q−1 dt.
Dividing both sides by Γ(p + q) 6= 0 yields the desired.
Application 3: Integration
We can use this to compute ∫ 1
0
dx√1− x1/4
.
Indeed, making the u-substitution u = x1/4 (so that dx = 4u3 du) yields∫ 1
0
dx√1− x1/4
=
∫ 1
0
4u3(1− u)−1/2 du =4Γ(4)Γ( 1
2 )
Γ( 92 )
.
But we already know that Γ(4) = 3! = 6 and
Γ( 92 ) = 7
2 ·52 ·
32 ·
12 Γ( 1
2 ) = 10516 Γ( 1
2 ).
Therefore the above expression simplifies nicely to
4 · 6 · Γ( 12 )
10516 Γ( 1
2 )=
128
35.
Application 3: Integration
We can use this to compute ∫ 1
0
dx√1− x1/4
.
Indeed, making the u-substitution u = x1/4 (so that dx = 4u3 du) yields∫ 1
0
dx√1− x1/4
=
∫ 1
0
4u3(1− u)−1/2 du =4Γ(4)Γ( 1
2 )
Γ( 92 )
.
But we already know that Γ(4) = 3! = 6 and
Γ( 92 ) = 7
2 ·52 ·
32 ·
12 Γ( 1
2 ) = 10516 Γ( 1
2 ).
Therefore the above expression simplifies nicely to
4 · 6 · Γ( 12 )
10516 Γ( 1
2 )=
128
35.
Application 3: Integration
We can use this to compute ∫ 1
0
dx√1− x1/4
.
Indeed, making the u-substitution u = x1/4 (so that dx = 4u3 du) yields∫ 1
0
dx√1− x1/4
=
∫ 1
0
4u3(1− u)−1/2 du =4Γ(4)Γ( 1
2 )
Γ( 92 )
.
But we already know that Γ(4) = 3! = 6 and
Γ( 92 ) = 7
2 ·52 ·
32 ·
12 Γ( 1
2 ) = 10516 Γ( 1
2 ).
Therefore the above expression simplifies nicely to
4 · 6 · Γ( 12 )
10516 Γ( 1
2 )=
128
35.
Application 3: Integration
Let’s do a wackier example and compute∫ 1
0
dx3√x2(1− x)
.
In the same vein as before, this rewrites as∫ 1
0
dx3√x2(1− x)
=
∫ 1
0
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√
3.
Hence the desired integral equals 2π√3
.
Application 3: Integration
Let’s do a wackier example and compute∫ 1
0
dx3√x2(1− x)
.
In the same vein as before, this rewrites as∫ 1
0
dx3√x2(1− x)
=
∫ 1
0
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√
3.
Hence the desired integral equals 2π√3
.
Application 3: Integration
Let’s do a wackier example and compute∫ 1
0
dx3√x2(1− x)
.
In the same vein as before, this rewrites as∫ 1
0
dx3√x2(1− x)
=
∫ 1
0
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√
3.
Hence the desired integral equals 2π√3
.
Application 3: Integration
Let’s do a wackier example and compute∫ 1
0
dx3√x2(1− x)
.
In the same vein as before, this rewrites as∫ 1
0
dx3√x2(1− x)
=
∫ 1
0
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√
3.
Hence the desired integral equals 2π√3
.
Application 4: Sums
Recall that the binomial coefficient(nk
)is computed via(
n
k
):=
n!
(n − k)!k!.
Of interest to us are the central binomial coefficients(
2nn
).
In combinatorics, power series are very useful tools for either solving orestimating the terms of recursions. One useful power series is
∞∑n=0
(2n
n
)zn =
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
Recall that the binomial coefficient(nk
)is computed via(
n
k
):=
n!
(n − k)!k!.
Of interest to us are the central binomial coefficients(
2nn
).
In combinatorics, power series are very useful tools for either solving orestimating the terms of recursions. One useful power series is
∞∑n=0
(2n
n
)zn =
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
Recall that the binomial coefficient(nk
)is computed via(
n
k
):=
n!
(n − k)!k!.
Of interest to us are the central binomial coefficients(
2nn
).
In combinatorics, power series are very useful tools for either solving orestimating the terms of recursions. One useful power series is
∞∑n=0
(2n
n
)zn =
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
To motivate the following discussion, write
1(2nn
) =1
(2n)!n!n!
=n!n!
(2n)!.
The form of this fraction highly suggests we can write it as a Betaintegral. However, this doesn’t quite work, as
n!n!
(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
To motivate the following discussion, write
1(2nn
) =1
(2n)!n!n!
=n!n!
(2n)!.
The form of this fraction highly suggests we can write it as a Betaintegral. However, this doesn’t quite work, as
n!n!
(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
To motivate the following discussion, write
1(2nn
) =1
(2n)!n!n!
=n!n!
(2n)!.
The form of this fraction highly suggests we can write it as a Betaintegral. However, this doesn’t quite work, as
n!n!
(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
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.
Now
(n − 1)!n!
(2n)!zn =
Γ(n)Γ(n + 1)
Γ(2n + 1)zn
=
∫ 1
0
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.
Now
(n − 1)!n!
(2n)!zn =
Γ(n)Γ(n + 1)
Γ(2n + 1)zn
=
∫ 1
0
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
∫ 1
0
tn−1(z(1− t))n dt
=
∫ 1
0
∞∑n=1
tn−1(z(1− t))n dt
=
∫ 1
0
z(1− t)
1− tz(1− t)dt =
∫ 1
0
1− t
t2 − t + 1z
dt.
A bit of elbow grease (try it; it’s not as scary as it looks!) shows that this
integral equals√
z4−z arcsin(
√z
2 ).
Application 4: Sums
Differentiating both sides with respect to z yields
∞∑n=1
n!n!
(2n)!zn = F ′(z) =
z
4− z+
4√z
(4− z)3/2arcsin
(√z
2
).
Hence the value of the sum is 1 + F ′(z). In particular,
∞∑n=0
1(2nn
) = 1 + F ′(1) = 1 +1
3+
4
33/2· π
6=
36 + 2π√
3
27.
Application 4: Sums
Differentiating both sides with respect to z yields
∞∑n=1
n!n!
(2n)!zn = F ′(z) =
z
4− z+
4√z
(4− z)3/2arcsin
(√z
2
).
Hence the value of the sum is 1 + F ′(z). In particular,
∞∑n=0
1(2nn
) = 1 + F ′(1) = 1 +1
3+
4
33/2· π
6=
36 + 2π√
3
27.
Some Extra Problems
Problem 1 (Classic?)
Let Cn := 1n+1
(2nn
)be the nth Catalan number. Prove that
∞∑n=0
1
Cn= 2 +
4√
3π
27.
Problem 2 (AMM 11103, Gregory Galperin and Hillel Gauchman)
Prove that for every positive integer n,
n∑k=1
1
k(nk
) =1
2n−1
n∑k=1k odd
(nk
)k.