# Calculate the Gauss integral without squaring it first

We know that the integral $$I = \int_{-\infty}^{\infty} \mathrm{d}x e^{-x^2}$$ can be calculated by first squaring it and then treat it as a $2-$dimensional integral in the plane and integrate it in polar coordinates.

Are there any other ways to calculate it? I know that we may use the relation $$\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin{\pi x}},$$ but this, in effect, is still taking the square.

Well, after I write down the above text, I figure that maybe there is no way to calculate it without squaring, since, after all, the result contains a square root, and it seems no elementary function can "naturally" produce a square root of $\pi$ starting from natural numbers (though I don't know how to describe this more concretely; you are also welcome to comment on this point). Nevertheless I still post this question in case there are some other ideas.

EDIT: the Fourier transformation method at Computing the Gaussian integral with Fourier methods? appears kind of cheat to me, since the very proof of the Fourier transformation formula actually makes use of the value of the Gauss integral (at least in this wiki page http://en.wikipedia.org/wiki/Fourier_inversion_theorem#Proof).

Thank you.

• I'm interested as well. This would be useful in probability if there were another way to compute the integral. – Clarinetist Jun 25 '14 at 4:38
• There are several other ways. See, for instance, math.uconn.edu/~kconrad/blurbs/analysis/gaussianintegral.pdf. – KCd Jun 25 '14 at 4:42
• This question has been asked before. See math.stackexchange.com/questions/149824/… – KCd Jun 25 '14 at 4:45
• See my answer to this question, towards the bottom of the page. – Lucian Jun 25 '14 at 6:24
• No offense, @Lucian, but you rely on $(1/2)!^2=\pi$, which is assuming the OP's result is true. – Ron Gordon Jun 25 '14 at 21:17

I will cover some methods that can be used that do not use squaring (at least not to start with) and add a link to the paper below:

1) Asymptotic Method: It is possible to show that \begin{equation*} 1+x\leq e^x\leq\frac{1}{1-x} \end{equation*} using power series expansions. Replace $x$ with $x^2$, raise all the terms to a positive power $n$, and integrate from $0$ to $1$. The next step involves the following change of variables for each of the terms from left to right: $x=\sin(\theta)$, $x=\frac{y}{\sqrt{n}}$, $x=\tan(\theta)$. Then we have \begin{equation*} \sqrt{n}I_{2n+1}\leq\int^{\sqrt{n}}_{0}e^{-y^2}dy\leq \sqrt{n}I_{2n-2} \end{equation*} where $I=\int^{\infty}_{-\infty}e^{-x^2/2}$. Setting $I_0,I_1$ and $I_k$ as specific values, we have a complicated use of the squeeze theorem and limits to establish the result.

2) Original Proof: This was an attempt by Laplace to compute the integral \begin{equation*} \int^1_0\frac{dx}{\sqrt{-\log (x)}}. \end{equation*} It involved the substitution $y=\sqrt{-\log(x)}$ and Euler's formula \begin{equation*} \int^1_0 \frac{x^r dx}{\sqrt{1-x^{2s}}}\int^1_0 \frac{x^{s+r}dx}{\sqrt{1-x^{2s}}}=\frac{1}{s(r+1)}\frac{\pi}{2}. \end{equation*}

3) Contour Integration: The function $e^{-z^2/2}$ has no poles, so this approach involves introduction of the function \begin{equation*} \frac{e^{-z^2/2}}{1-e^{-\sqrt{\pi}(1+i)z}} \end{equation*} and a rectangular contour.

4) Using Stirling's formula: This approach is done by clever manipulation of Stirling's formula \begin{equation*} n!\sim \frac{n^n}{e^n}\sqrt{2\pi n}~\text{as}~n\to \infty \end{equation*} and De Moivre's asymptotic formula \begin{equation*} n!\sim C(\frac{n}{e})^n\sqrt{n} \end{equation*} to get the result.

• The proof shows that determining the constant $C$ in Stirling's formula is $\sqrt{2\pi}$ is equivalent to the result. A recurrence relation derived in the asymptotic approach is then used to show this, and De Moivre's asymptotic formula is used as part of the manipulation. – user230715 Jun 3 '15 at 18:20
We will show directly that for real $x\geq 0$ $$1-x^2 \leq e^{-x^2} \leq \frac{1}{1+x^2}$$ Let $a(x) = 1-x^2$ and $b(x)= e^{-x^2}$; then $a(0) = b(0) = 1$. And for $x\geq 0$, $$\frac{da(x)}{dx} = -2x\\ \frac{db(x)}{dx} = -2xe-{x^2}$$ Since $e-{x^2}$ is positive and less than or equal to $1$, and $-2x \leq 0$, this shows that $$\frac{da(x)}dx = \leq \frac{db(x)}dx$$ so $a(x)$ and $b(x)$ are equal at $x=0$ and for all positive $x$ the derivative of $a(x)$ is not greater than the derivative of $b(x)$ so $a(x)$ is never greater than $b(x)$: $$1-x^2 \leq e^{-x^2}$$ Now we wish to show $e^{-x^2} \leq \frac{1}{1+x^2}$; since both are positive we will show the equivalent relation obtained by taking reciprocals $e^{x^2} \geq 1+x^2$:
Let $c(x) = \frac{1}{1+x^2}$ and $e(x) = e^{x^2}$; again we have $c(0) = e(0) = 1$. $$\frac{dc(x)}{dx} = 2x\\ \frac{de(x)}{dx} = 2xe{x^2} \\ e^{x^2} \geq 1 \implies \frac{de(x)}{dx} \geq \frac{dc(x)}{dx}$$ so everywhere for $x\geq 0$ this shows that $$e^{x^2} \geq 1+x^2$$ which then implies $$e^{-x^2} \leq \frac{1}{1+x^2}$$ Comining these two results we have proven directly that $$1-x^2 \leq e^{-x^2} \leq \frac{1}{1+x^2}$$