How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates?

How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?

Solution 1:

Method 1. Let $I$ denote the integral. Then for $s > 0$, the substitution $t = x\sqrt{s}$ gives

$$ \int_{0}^{\infty} e^{-sx^2} \; dx = \frac{1}{\sqrt{s}} \int_{0}^{\infty} e^{-t^2} \; dt = \frac{I}{\sqrt{s}}. $$

Thus for $u = x^2$,

$$\begin{align*}I^2 &= \int_{0}^{\infty} I e^{-x^2} \; dx = \int_{0}^{\infty} \frac{I}{2\sqrt{u}}e^{-u} \; du = \int_{0}^{\infty} \frac{1}{2}e^{-u}\int_{0}^{\infty}e^{-ut^2}\;dtdu\\ &= \int_{0}^{\infty}\int_{0}^{\infty} \frac{1}{2}e^{-u}e^{-ut^2}\;dudt = \int_{0}^{\infty}\frac{1}{2(t^2+1)}\;dt = \frac{\pi}{4}, \end{align*}$$

which proves the desired result. Note that this is in fact equivalent to the proof in Wikipedia.

Method 2. We assume that we are aware of the Wallis product

$$ \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}. $$

Now, it is not hard to find that

$$ f_n(x) = \begin{cases} \left(1 - \tfrac{x^2}{n}\right)^{n} & 0 \leq x \leq \sqrt{n} \\ 0 & \text{otherwise}. \end{cases} $$

increases to $e^{-x^2}$. (Simple application of Bernoulli's inequality will suffice.) Then by monotone convergence theorem,

$$ I := \int_{0}^{\infty} e^{-x^2} \; dx = \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left(1 - \frac{x^2}{n}\right)^{n} \; dx = \lim_{n\to\infty} \sqrt{n} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \; d\theta, $$

where we have used the substitution $x = \sqrt{n}\,\cos\theta$. Now let

$$I_n = \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \; d\theta. $$

Then for $n \geq 1$,

$$\begin{align*}I_n &= \left[\sin^{2n}\theta (-\cos\theta)\right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2n \sin^{2n-1}\theta \cos\theta (-\cos\theta) \; d\theta \\ &= \int_{0}^{\frac{\pi}{2}} 2n \sin^{2n-1}\theta (1-\sin^2 \theta) \; d\theta = 2n(I_{n-1} - I_n) \end{align*}$$

and we have $I_0 = 1$ and $I_n = \frac{2n}{2n+1} I_{n-1}$. Thus

$$ I_n = \prod_{k=1}^{n} \frac{2k}{2k+1}. $$

Therefore, by Wallis product,

$$ I^2 = \lim_{n\to\infty} n \prod_{k=1}^{n} \left( \frac{2k}{2k+1} \cdot \frac{2k}{2k+1} \right) = \lim_{n\to\infty} \frac{n}{2n+1} \prod_{k=1}^{n} \left( \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \right) = \frac{1}{2} \cdot \frac{\pi}{2} $$

and we have the desired result. (Conversely, knowing $I$, it can be used to prove the Wallis product formula.)

Solution 2:

Because $x \mapsto e^x$ is convex, $\displaystyle \left( 1- \frac{u}{n} \right)^n \leq e^{-u} \leq \left( 1+ \frac{u}{n} \right)^{-n}$, so $\displaystyle \int_0^{\sqrt{n}} \left( 1- \frac{x^2}{n} \right)^n dx \leq \int_0^{\sqrt{n}} e^{-x^2} dx \leq \int_0^{\sqrt{n}} \left( 1+ \frac{x^2}{n} \right)^{-n} dx$.

We use the change of variables $x= \sqrt{n} \sin(\theta)$: $\displaystyle \int_0^{\sqrt{n}} \left( 1- \frac{x^2}{n} \right)^n dx = \int_0^{\pi/2} \cos^{2n+1} (\theta) d\theta= \sqrt{n} W_{2n+1}$, where $W_{2n+1}$ is a Wallis integral.

If $p \geq \sqrt{n}$, $\displaystyle \int_0^{\sqrt{n}} \left(1+ \frac{x^2}{n} \right)^{-n} dx \geq \int_0^p \left(1+ \frac{x^2}{n} \right)^{-n} dx$. We use the change of variables $x=\sqrt{n} \tan (\theta)$ to find: $\displaystyle \int_0^{\arctan(p)} \sqrt{n} \cos^{2n+2} (\theta) d \theta \underset{p \to + \infty}{\longrightarrow} \sqrt{n} W_{2n+2}$.

Finally, $\displaystyle \sqrt{n} W_{2n+1} \leq \int_0^{\sqrt{n}} e^{-x^2}dx \leq \sqrt{n} W_{2n+2}$. But $\displaystyle W_n \underset{n \to + \infty}{\sim} \sqrt{\frac{\pi}{2n}}$ so $\displaystyle \int_0^{+ \infty} e^{-x^2}dx= \frac{\sqrt{\pi}}{2}$.

With this method, an asymptotic development of $W_n$ gives an asymptotic development of the integral.