gamma derivation

There are few creatures in mathematics as interesting yet, somehow, useful as the Gamma function,

An interesting result about this function is that $\Gamma (n)=(n-1)!$ for positive integers $n\in \mathbb{N}$. But why is that so? Better yet, why is the gamma function defined so, how is it related to factorials, and what does its values at non-integers mean?

First, consider the integral

It is not a polynomial function, so we cannot use the power rule. However, if you remember your calculus well, you may recall integration by parts:

Write $u=\ln x$ and $\mathrm{d}v=\mathrm{d}x$. Then $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$ and $v=x$. So

What happens if we have higher powers of $\ln x$?

Substitute $u={\ln }^{2}x$ and $\mathrm{d}v=\mathrm{d}x$. Then $\mathrm{d}u=\frac{2\ln x}{x}\mathrm{d}x$ and $v=x$.

Using integration by parts;

Interestingly, we get $I(x)$ as part of the integration result of $I_2(x)$. If we fully integrate the result, we get:

What happens with the third power? Take a guess, you're probably right.

Take $u={\ln }^{3}x$ and $\mathrm{d}v=\mathrm{d}x$. Then $\mathrm{d}u=\frac{3{\ln }^{2}x}{x}\mathrm{d}x$ and $v=x$.

Again, we have $I_2(x)$ as part of the integration result of $I_3(x)$. Evaluating the integral fully gives:

Indeed, it always occurs that $I_n(x)=x{\ln }^{n}x-n{I}_{n-1}(x)$. But why is this a useful result?

Take a look at the coefficients of the last term, $x$.

Those are factorials, with alternating signs (positive, negative).

Our goal is two-fold:

• To get rid of the leading terms involving $\ln x$.
• To get rid of the alternation in signs.

The leading terms have one thing in common: they are all a product of $x$ and a power of $\ln x$. Thus, they all tend to zero at both $x=1$ (since $\ln 1=0$) and at $x=0$. We need to retain the last term with just $x$, which we can do by evaluating the integral over the interval $(0,1)$.

To do this, we need to negate the result for odd $n$. This is easily achieved by dividing the result by $(-1{)}^n$, giving the integral

But the integral now looks complicated. Can we simplify it?

Substitute $u=\ln \frac{1}{x}$, so that:

Note that as $x\to 0^{+}$, $\frac{1}{x}\to {\infty }^{+}$, therefore $u\to {\infty }^{+}$, and when $x\to 1$, $\frac{1}{x}\to 1$, so $u\to 0$.

Thus;

This is the exact form of the gamma function, evaluated at $s=n+1$.

The gamma function has various useful properties. For example, although it collapses to the factorial function at positive integers, it is well-defined for all complex numbers except zero and the non-negative integers. It is sometimes used as an analytic continuation of the factorial function to fractional and complex values. Over non-integers, the logarithmic terms are consequential in the value giving rise to the gamma extensions.

As an ode to its relevance, $\Gamma$ appears in multiple areas of mathematics, including:

• Complex analysis (definition of the Riemann zeta function):

• Probability theory (definition of the beta distribution):

• Number theory (the functional equation of the Riemann zeta function):

• And many more, left as an exercise to the reader.