Named after French mathematician Augustin-Louis Cauchy, Project Cauchy column is where I invite some of the HFI Programming club members to provide neat proofs or explanations about some number theory puzzles. I highly suggest that you read these articles with a pencil and paper so you can sketch things out and scribble solutions to exercises as you come across them. This week, Travor is demonstrateing us one of the most commonly used extension of the factorial function to complex numbers.
It is well-known that factorial is defined by the following recursive relation,
n!=n(n−1)!
with 0!=1, but however it is possible to generalize this operation to complex numbers. Let's begin our journey of generalizations!
First generalization: integral representation
It can be easily shown that the following integral satisfies
∫0∞e−λtdt=λ1
Differentiation with respect to s on both side for n−1 times gives
∫0∞tn−1e−λtdt=λn(n−1)!
Setting λ=1 gives us the first generalization of factorial: the Gamma function.
(s−1)!=Γ(s)=∫0∞ts−1e−tdt
In fact, by integration by parts we can show that the Gamma function satisfies the recursive relationship:
Γ(s+1)=sΓ(s)(1)
Furthermore, we have the relation
∫0∞ts−1e−λtdt=λsΓ(s)
Digression: the Beta function
To convenience our derivation process, let's define the Euler integral of the first kind: the Beta function B(x,y):
B(x,y)=∫01τx−1(1−τ)y−1dτ(2)
which can be seen as a convolution between two power functions:
B(x,y;t)=∫0tτx−1(t−τ)y−1dτ
If we were to perform Laplace transform on both side, we get
L{B(x,y;t)}(λ)=λx+yΓ(x)Γ(y)
If we juxtapose this fact with the relation
L{tx+y−1}(λ)=λx+yΓ(x+y)
then we see that
B(x,y)=B(x,y;1)=Γ(x+y)Γ(x)Γ(y)
which will be extremely useful for us to generalize factorial even further.
Γ(s) evaluated at complex numbers
Via taking absolute variable, we know that the Gamma integral converges whenever ℜ(s)>0
∣Γ(s)∣≤∫0∞tℜ(s)−1e−tdt
which means that this improper integral is converges absolutely on the right half plane. Hence, a stronger definition is needed for us to expand it to the entire complex plane.
Γ(s) as a limit
Let's consider a sequence of functions:
fn(t,s)={ts−1(1−nt)n00≤x≤nx>n
Then by the exponential limit we see that
fn(t,s)→ts−1(1−nt)n
in a pointwise sense. However, it is possible to show that this sequence converges uniformly for t∈[0,+∞). Let's set T>0 such that for all t>T we have
By the uniform continuity of e−t on [0,T], all we need is to prove that bn(t) converges uniformly to t. In fact, for all n>t we can use the Taylor expansion of logarithm to obtain
As a result, we conclude that fn(s,t) converges uniformly to ts−1e−t, which allows us to interchange the limit operation and integral to obtain
n→∞lim∫0nts−1(1−nt)ndt=Γ(s)(3)
In the following procedure, we are going to expand the left hand side limit in a subtle sense so that the right hand side can be analytically continued to the left half plane.
As a result, the product converges absolutely for all s∈C, giving us the Weierstrass product representation of Gamma function:
Γ(s)1=seγsk=1∏∞(1+ks)e−s/k
which allows us to analytically continue Γ(s) to the entire complex planes except for nonpositive integers:
Γ(s)=se−γsk=1∏∞(1+ks)−1es/k(4)
Remark: (4) also reveals Γ(s) is non-zero for all s∈C
Now, we successfully expanded factorial to the entire complex plane except at negative integers, but Gamma function has some other brilliant properties. Let's have a look at some of them:
Alternative Definition for Euler-Mascheroni constant
From the last article, we know that Euler-Mascheroni constant is defined by
γ=n→∞lim(Hn−logn)=1−∫1∞x2{x}dx
However, it is possible for us to create a new definition for γ by using Γ(s).
Remark: I hypothesize this explains why they name the function Γ and the constant γ since they are highly correlated.
To begin with, we take logarithm on (4) to get
logΓ(s)=−logs−γs+k=1∑∞{ks−log(1+ks)}
If we were to define Digamma functionψ(s) as the logarithmic derivative of Γ(s), then
ψ(s)=−γ−s1+k=1∑∞{k1−s+k1}
If we were to move the s1 term into the summation, we deduce the standard definition for Digamma function.
ψ(s)=−γ+m=0∑∞{m+11−m+s1}(5)
Plugging s=1 gives ψ(1)=Γ′(1)/Γ(1)=−γ. Because Γ(1)=1, we also know that Γ′(1)=−γ. Combining this with the original integral definition for Γ(s) gives this elegant integral identity to represent Euler-Mascheroni constant:
∫0∞e−tlog(t)dt=−γ(6)
Integral representation for Digamma function
While deriving (5) in the previous section, we introduce the Digamma function, so why don't we do some calculus on that
Combining all these gives us the ultimate integral definition for ψ(s):
ψ(s)=∫01{loglog(x1)+x−1xs−1−1}dx(7)
Conclusion
In this blog, we first use the technique of differentiation under integral to deduce an integral representation for factorial that introduces the concept of Gamma function Γ(s). Then, by using an identity connecting B(x,y) and Γ(s), we obtain a product formula that turns Γ(s) into a meromorphic function on C. Using this newly obtained product formula, we are also able to discover some new identities. In fact, Gamma function is a function that often appears in the field of analytic number theory, and we will begin future investigation based on the following identity (which you could try prove it yourself):