# Recursion on Integral

Factorial is a simple straightforward function. It’s the multiplication of integers from $1$ to $n$.

```
factorial n = product [1..n]
```

That is, by observation, it can be written as a recursion.

```
factorial 1 = 1
factorial n = n * factorial (n - 1)
```

We may add another rule: $0! = 1$, for convenient when writing binomial expansion. Since we do that a lot in probability.

But it still leaves a problem. That is everything we consider is discrete math. So it is impossible for $0.5!$ to be calculated.

Well, if we take a look at usage (mostly related to probability), it might not seems to be a problem at all. But for someone study pure math, there should be something that can answer this question.

Luckily (?) that we have a number $e$, with a strange property

\[e^x = \frac{d}{dx} e^x.\]It just like the fixed-point combinator that I’m talk in previous posts. Just this time we change function call into diff-integral. With the terminate point at

\[\begin{align} \int_0^\infty \frac{1}{e^x} \;dx &= - \frac{1}{e^\infty} + \frac{1}{e^0} \\ &= 1. \end{align}\]That’s allow us to write the recursion using integral, like this

\[\Gamma(n) = \int_0^\infty \frac{1}{e^t} t^{n-1} \;dt.\]We may inspect its property by taking integral

\[\begin{align} \Gamma(n) &= \int_0^\infty \frac{1}{e^t} t^{n-1} \;dt \\ &= \int_0^\infty u \;dv & (\; \text{let}\; u = \frac{1}{e^t}, dv = t^{n-1} \;dt \;) \\ &= \left[ uv \right]_0^\infty - \int_0^\infty v \;du \\ &= \frac{1}{n} \left( \lim_{t \to \infty} \frac{1}{e^t} t^n - \frac{1}{e^0} 0^n \right) - \int_0^\infty v \;du \\ &= 0 - \int_0^\infty v du \\ &= \frac{1}{n} \int_0^\infty \frac{1}{e^t} t^n \;dt \\ &= \frac{\Gamma(n+1)}{n}. \end{align}\]Thus we see

\[\begin{align} \Gamma(1) &= 1 \\ \Gamma(n) &= (n-1) \Gamma(n-1) \\ &= (n-1)(n-2) \Gamma(n-2) \\ &= \cdots \end{align}\]Therefore

\[n! = \Gamma(n+1).\]Now we equipped with the knowledge for $0.5!$ calculation. That is

\[\begin{align} \Gamma(1.5) &= \frac{1}{2} \Gamma\left(\frac{1}{2}\right), \\ \Gamma(0.5) &= \int_0^\infty \frac{1}{e^t} \frac{1}{\sqrt t} \;dt \\ &= 2 \int_0^\infty \frac{1}{ e^{u^2} } \;du & (\; \text{let}\; du = \frac{1}{\sqrt t} \;) \\ &= \int_{-\infty}^\infty \frac{1}{ e^{u^2} } \;du \\ &= \sqrt\pi, \\ \Gamma(1.5) &= \frac{\sqrt\pi}{2}. \end{align}\]When we try to find $\Gamma(0.5)$, we need double integral and the change of coordinate. Anyone who’s interest can find the detail in the following video.

The good thing that we describe it in term of integral is that, if we find it’s too hard to analyse, we may use Riemann integral to approximate the result instead.

*author*