Factorial, Gamma Function, and More

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

$\int_0^\infty e^{-\lambda t}\mathrm dt=\frac1\lambda$

Differentiation with respect to $s$ on both side for $n-1$ times gives

$\int_0^\infty t^{n-1}e^{-\lambda t}\mathrm dt={(n-1)!\over\lambda^n}$

Setting $\lambda=1$ gives us the first generalization of factorial: the Gamma function.

$(s-1)!=\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}\mathrm dt$

In fact, by integration by parts we can show that the Gamma function satisfies the recursive relationship:

$\Gamma(s+1)=s\Gamma(s)\tag1$

Furthermore, we have the relation

$\int_0^\infty t^{s-1}e^{-\lambda t}\mathrm dt={\Gamma(s)\over\lambda^s}$

Digression: the Beta function

To convenience our derivation process, let's define the Euler integral of the first kind: the Beta function $B(x,y)$:

$B(x,y)=\int_0^1\tau^{x-1}(1-\tau)^{y-1}\mathrm d\tau\tag2$

which can be seen as a convolution between two power functions:

$B(x,y;t)=\int_0^t\tau^{x-1}(t-\tau)^{y-1}\mathrm d\tau$

If we were to perform Laplace transform on both side, we get

$\mathscr L\{B(x,y;t)\}(\lambda)={\Gamma(x)\Gamma(y)\over\lambda^{x+y}}$

If we juxtapose this fact with the relation

$\mathscr L\{t^{x+y-1}\}(\lambda)={\Gamma(x+y)\over\lambda^{x+y}}$

then we see that

$B(x,y)=B(x,y;1)={\Gamma(x)\Gamma(y)\over\Gamma(x+y)}$

which will be extremely useful for us to generalize factorial even further.

$\Gamma(s)$ evaluated at complex numbers

Via taking absolute variable, we know that the Gamma integral converges whenever $\Re(s)>0$

$|\Gamma(s)|\le\int_0^\infty t^{\Re(s)-1}e^{-t}\mathrm dt$

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.

$\Gamma(s)$ as a limit

Let's consider a sequence of functions:

$f_n(t,s)= \begin{cases} t^{s-1}\left(1-\frac tn\right)^n & 0\le x\le n \\ 0 & x>n \end{cases}$

Then by the exponential limit we see that

$f_n(t,s)\to t^{s-1}\left(1-\frac tn\right)^n$

in a pointwise sense. However, it is possible to show that this sequence converges uniformly for $t\in[0,+\infty)$. Let's set $T>0$ such that for all $t>T$ we have

$|t^{s-1}e^{-t}|<\varepsilon$

Then, let's consider the interval $[0,T]$:

$\begin{aligned} |t^{s-1}e^{-t}-f_n(t,s)| &=t^{\Re(s)-1}\left|e^{-t}-\left(1-\frac tn\right)^n\right| \\ &=t^{\Re(s)-1}\left|e^{-t}-e^{-b_n(t)}\right| \end{aligned}$

where we define $b_n(t)$ as

$b_n(t)=-n\log\left(1-\frac tn\right)$

By the uniform continuity of $e^{-t}$ on $[0,T]$, all we need is to prove that $b_n(t)$ converges uniformly to $t$. In fact, for all $n>t$ we can use the Taylor expansion of logarithm to obtain

$\begin{aligned} |b_n(t)-t| &=\left|-n\log\left(1-\frac tn\right)-t\right| \\ &=\left|-n\left[\frac tn+\mathcal O\left(1\over n^2\right)\right]-t\right| \\ &=\mathcal O\left(\frac1n\right) \end{aligned}$

As a result, we conclude that $f_n(s,t)$ converges uniformly to $t^{s-1}e^{-t}$, which allows us to interchange the limit operation and integral to obtain

$\lim_{n\to\infty}\int_0^n t^{s-1}\left(1-\frac tn\right)^n\mathrm dt=\Gamma(s)\tag3$

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.

Expanding the integral sequence

Performing a change of variables on (3) gives

$\Gamma_n(s)\triangleq\underbrace{\int_0^nt^{s-1}\left(1-\frac tn\right)^n\mathrm dt}_{t=nr}=n^s\int_0^1r^{s-1}(1-r)^n\mathrm dr$

In fact, the right hand side can be expressed by Beta function as in (3), which yields

$\Gamma_n(s)=n^sB(s,n+1)=n^s{\Gamma(s)\Gamma(n+1)\over\Gamma(s+n+1)}$

Continuous application of (1) gives us

$\begin{aligned} \Gamma_n(s) &={\Gamma(s)n^sn!\over\Gamma(s+n+1)} \\ &={n^sn!\over s(s+1)(s+2)\cdots(s+n)} \\ &={n^s\over s}\prod_{k=1}^n{k\over s+k} \\ \end{aligned}$

which transforms Gamma function into a product representation, however this expression still looks ugly, why not go deeper?

Weierstrass product for $\Gamma(s)$

First, let's turn this equation up side down to obtain

$\begin{aligned} {1\over\Gamma_n(s)} &=sn^{-s}\prod_{k=1}^n\left(1+\frac sk\right) \\ &=se^{-s\log n}\prod_{k=1}^n\left(1+\frac sk\right) \end{aligned}$

Then, employing the fact that $H_n\triangleq\sum_{k=1}^n\frac1k=\log n+\gamma+\mathcal O(1/n)$ we can replace the logarithm with harmonic numbers:

$\begin{aligned} {1\over\Gamma_n(s)} &=se^{-s[H_n-\gamma+\mathcal O(1/n)]}\prod_{k=1}^n\left(1+\frac sk\right) \\ &=se^{s\gamma+\mathcal O(1/n)}\prod_{k=1}^n\left(1+\frac sk\right)e^{-s/k} \end{aligned}$

Now, if we take logarithm on the product, we have

$\begin{aligned} \log\left|\prod_{k=1}^n\left(1+\frac sk\right)e^{-s/k}\right| &\le|s^2|\sum_{k=1}^n{1\over k^2}<{|s|^2\pi^2\over6} \end{aligned}$

As a result, the product converges absolutely for all $s\in\mathbb C$, giving us the Weierstrass product representation of Gamma function:

${1\over\Gamma(s)}=se^{\gamma s}\prod_{k=1}^\infty\left(1+\frac sk\right)e^{-s/k}$

which allows us to analytically continue $\Gamma(s)$ to the entire complex planes except for nonpositive integers:

$\Gamma(s)={e^{-\gamma s}\over s}\prod_{k=1}^\infty\left(1+\frac sk\right)^{-1}e^{s/k}\tag4$

Remark: (4) also reveals $\Gamma(s)$ is non-zero for all $s\in\mathbb C$

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

$\gamma=\lim_{n\to\infty}(H_n-\log n)=1-\int_1^\infty{\{x\}\over x^2}\mathrm dx$

However, it is possible for us to create a new definition for $\gamma$ by using $\Gamma(s)$.

Remark: I hypothesize this explains why they name the function $\Gamma$ and the constant $\gamma$ since they are highly correlated.

To begin with, we take logarithm on (4) to get

$\log\Gamma(s)=-\log s-\gamma s+\sum_{k=1}^\infty\left\{\frac sk-\log\left(1+\frac sk\right)\right\}$

If we were to define Digamma function $\psi(s)$ as the logarithmic derivative of $\Gamma(s)$, then

$\psi(s)=-\gamma-\frac1s+\sum_{k=1}^\infty\left\{\frac1k-{1\over s+k}\right\}$

If we were to move the $\frac1s$ term into the summation, we deduce the standard definition for Digamma function.

$\psi(s)=-\gamma+\sum_{m=0}^\infty\left\{{1\over m+1}-{1\over m+s}\right\}\tag5$

Plugging $s=1$ gives $\psi(1)=\Gamma'(1)/\Gamma(1)=-\gamma$. Because $\Gamma(1)=1$, we also know that $\Gamma'(1)=-\gamma$. Combining this with the original integral definition for $\Gamma(s)$ gives this elegant integral identity to represent Euler-Mascheroni constant:

$\int_0^\infty e^{-t}\log(t)\mathrm dt=-\gamma\tag6$

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

$\begin{aligned} \psi(s) &=-\gamma+\sum_{m=0}^\infty\int_0^1(x^m-x^{m+s-1})\mathrm dt \\ &=-\gamma+\int_0^1(1-x^{s-1})\sum_{m=0}^\infty x^m\mathrm dt \\ &=-\gamma+\int_0^1{1-x^{s-1}\over1-x}\mathrm dt \end{aligned}$

Well, the constant lying outside is not beautiful, so why not use (6):

$\begin{aligned} -\gamma &=\underbrace{\int_0^\infty e^{-t}\log(t)\mathrm dt}_{x=e^{-t}} \\ &=-\int_1^0\log(-\log x)\mathrm dx \\ &=\int_0^1\log\log\left(\frac1x\right)\mathrm dx \end{aligned}$

Combining all these gives us the ultimate integral definition for $\psi(s)$:

$\psi(s)=\int_0^1\left\{\log\log\left(\frac1x\right)+{x^{s-1}-1\over x-1}\right\}\mathrm dx\tag7$

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 $\Gamma(s)$. Then, by using an identity connecting $B(x,y)$ and $\Gamma(s)$, we obtain a product formula that turns $\Gamma(s)$ into a meromorphic function on $\mathbb C$. 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):

$\sum_{n=1}^\infty{1\over n^s}={1\over\Gamma(s)}\int_0^\infty{x^{s-1}\over e^x-1}\mathrm dx$