45  Special integrals*

There are many reasons to learn integrals. But the most compelling reason is that math is no longer the same with integrals. We can have many amazing results with integrals that were otherwise not imaginable. This section introduces two integrals that are of special importance to continuous distributions.

Proposition 45.1 (Guassian integral) \[\int_{-\infty}^{+\infty}e^{-x^{2}}dx=\sqrt{\pi}\]

This is known as Gaussian integral, which is the kernel of the PDF of the normal distribution. It also amazingly relates two of the most famous constants in mathematics. It is not integrable by normal integration techniques. But it can be proved by switching to the polar coordinate.

Proof. Let \(I=\int_{-\infty}^{+\infty}e^{-x^{2}}dx\). \[\begin{aligned} I^2= & \int_{-\infty}^{+\infty}e^{-x^{2}}dx\int_{-\infty}^{+\infty}e^{-y^{2}}dy\\ = & \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^{2}+y^{2})}dxdy\\ = & \int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^{2}}rdrd\theta & dA=dxdy=rdrd\theta\\ = & \int_{0}^{2\pi}\int_{0}^{\infty}\frac{1}{2}e^{-u}dud\theta & \textrm{let }u=r^{2}\\ = & \frac{1}{2}\int_{0}^{2\pi}d\theta=\pi. \end{aligned}\] Thus, \(I=\sqrt{\pi}\).

Proposition 45.2 (Gamma function) \[\int_{0}^{\infty}t^{n}e^{-t}dt=n!\]

\(\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt\) is known as the Gamma function, which is definitely one of the most interesting functions in mathematics. It is the extension of factorials to real numbers or even complex numbers. It also has many interesting properties, such as \(\Gamma(n)=(n-1)!\), \(\Gamma(1/2)=\sqrt{\pi}\), \(\Gamma(3/2)=\sqrt{\pi}/2\), \(\Gamma'(1)=-\gamma\) and so on. The \((n-1)\) in the Gamma function is due to historical reasons and does not matter in our case. We will prove the integral with \(n\) instead of \((n-1)\).

Proof. There are many ways to prove this. One is to discover the recursive relationship \(\Gamma(n+1)=n\Gamma(n)\). But it does not give a clue why we need this integral to approximate the factorial. We start with an elementary integral

\[\int_{0}^{\infty}e^{at}dt=-\frac{1}{a}\] where \(a<0\). Differentiate both sides \(n\) times with respect to \(a\):

\[\begin{aligned}\int_{0}^{\infty}e^{at}tdt & = & -(-1)a^{-2}\\ \int_{0}^{\infty}e^{at}t^{2}dt & = & -(-1)(-2)a^{-3}\\ \int_{0}^{\infty}e^{at}t^{3}dt & = & -(-1)(-2)(-3)a^{-4}\\ & \vdots\\ \int_{0}^{\infty}e^{at}t^{n}dt & = & (-1)^{n+1}n!a^{-(n+1)} \end{aligned}\]

Let \(a=-1\), we have \[\int_{0}^{\infty}e^{t}t^{n}=n!\]