Proving $\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$

Question

How to prove $$\int_{0}^{+\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$$

Answer 1

This is an old favorite of mine.
Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$ Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$ $$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$ Now change to polar coordinates
$$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$ The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$ $$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$ So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.

Answer 2

A variation on Ross Millikan's answer.

We can start again with the observation $$\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-y^2}dy\right)=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)=V.$$ Now $V$ is simply the volume of the body $$-\infty < x,y < \infty,\qquad 0 < z < e^{-(x^2+y^2)},$$ or, equivalently, $$0 < x^2+y^2 < -\ln z,\qquad 0 < z < 1.$$ This implies that the body is a solid of revolution. Using the disk integration formula, we have $$V=\int_{0}^{1}\pi(-\ln z)dz=[\pi(z-z\ln z)]_{0}^1=\pi.$$

Answer 3

I know this:

Define $f$ and $g$ as:

$$f(x):=\left(\int_0^x e^{-t^2}dt\right)^{2} \ \ \ \text{and} \ \ \ g(x):=\left(\int_{0}^{1}\frac{e^{-x^{2}(t^{2}+1)}}{t^{2}+1}dt\right)$$

Now, $$f'(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$ and

$$g'(x)=\int_0^1 \frac{\partial}{dx}\left[\frac{e^{-x^2(t^2+1)}}{t^2+1}\right]dt = -2xe^{-x^{2}}\int_{0}^{1}e^{-x^{2}t^{2}}dt$$

So putting $t=tx$, get $\displaystyle\int_{0}^{1}e^{-x^{2}t^{2}}dt= \frac{1}{x}\displaystyle\int_{0}^{x}e^{-t^{2}}dt$

Then we get;

$$g'(x)=-2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$

Thus $f'(x)+g'(x)=0$ for all $x$, then $f(x)+g(x)$ is an constant function. Also $$f(0)+g(0)=\displaystyle\int_{0}^{1}\frac{1}{t^{2}+1}dt = \displaystyle\frac{\pi}{4}$$

Then $f(x)+g(x)=\displaystyle\frac{\pi}{4}$ for all $x$.

Now $\lim_{x \to{+}\infty}{g(x)}=0$

So $$\displaystyle\frac{\pi}{4} = \lim_{x \to{+}\infty}{f(x)+g(x)}=\lim_{x \to{+}\infty}{f(x)}= \left(\int_{0}^{\infty}e^{-t^{2}}dt\right)^{2}$$

Thus

$$\int_{0}^{\infty}e^{-t^{2}}dt=\sqrt{\frac{\pi}{4}}= \frac{\sqrt{\pi}}{2}$$

The end.

Answer 4

It might be worth mentioning that one also can use spherical coordinates in 3-dimensions analogously to the polar coordinates Ross Millikan used above: If $I$ denotes $\int_{-\infty}^{\infty} e^{-x^2}dx$, then we have $$I^3 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,dx\,dy\,dz$$ Switching to spherical coordinates this becomes $$\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty}\sin(\phi) \rho^2 e^{-\rho^2}\,d\rho\,d\phi\,d\theta$$ Doing the theta and $\phi$ integrations this becomes $$I^3 = 4\pi\int_0^{\infty}\rho^2 e^{-\rho^2}\,d\rho$$ One can then integrate parts in this, differentiating $\rho$ and integrating $\rho e^{-\rho^2}$. This leads us to $$I^3 = 2\pi \int_0^{\infty} e^{-\rho^2}\,d\rho$$ Note the right-hand side is exactly $2\pi * {I \over 2} = \pi I$. Thus $I^3 = \pi I$ and thus $I = \sqrt{\pi}$ as needed. Obviously polar coordinates are faster. Just sayin'...

Answer 5

An alternative derivation is to show that

$$\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; \mathrm{d}y=I,$$

where $I$ is your integral:

$$I:=\int_{0}^{\infty}e^{-x^2}\; \mathrm{d}x,$$

and then evaluate $I^2$ by reversing the order of integration. If $x>0$, then

$$\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; \mathrm{d}y=x\int_{0}^{\infty}e^{-{(xy)}^2}\; \mathrm{d}y=x\int_{0}^{\infty}e^{-u^2}\dfrac{\mathrm{d}u}{x}=\int_{0}^{\infty}e^{-u^2}\; \mathrm{d}u=I.$$

Thus

$$\begin{aligned}I^2&=\int_{0}^{\infty}e^{-x^2}\; \mathrm{d}x\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; \mathrm{d}y=\displaystyle\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}xe^{-x^2}e^{-x^{2}y^{2}}\; \mathrm{d}x\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}xe^{-x^{2}(1+y^2)}\; \mathrm{d}x=\int_{0}^{\infty}\mathrm{d}y\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty }\\ &=\int_{0}^{\infty }\dfrac{1}{2\left( 1+y^{2}\right) }\; \mathrm{d}y=\dfrac{1}{2}\left[ \arctan y\right] _{y=0}^{\infty }=\dfrac{\pi}{4}.\end{aligned}$$

So

$$I=\dfrac{\sqrt{\pi}}{2}.$$

Answer 6

Another proof, from G.M. Fichtengoltz, Calculus Course, page 612.

$$K=\int_{0}^{\infty} e^{-x^2} dx $$

It easy to see (and prove) that, $\max\{(1+t)e^{-t}\}=1$ at $t=0$, hence for all $t\in\mathbb{R}$:

$$(1+t)e^{-t}<1$$

Substitution of $t=\pm x^2$, leads us to:

$$(1-x^2)e^{x^2}<1 \ \ \ \ \text{and} \ \ \ \ \ (1+x^2)e^{-x^2}<1 $$

So,

$${1-x^2} <e^{-x^2}<\frac{1}{1+x^2} \ \ \ \ \ \ (x>0) $$

Now, at the left inequality we restrict our $x$ to be in $(0,1)$ (so that, $1-x^2>0$), and in the right inequality let $x>0$. Raising all the inequalities with natural number $n$, we get,

$$\underset{x\in (0,1)}{(1-x^2)^n<e^{-nx^2}} \ \ \ \ \text{and} \ \ \ \ \ \underset{x>0}{e^{-nx^2}<\frac{1}{(1+x^2)^n}}$$

Integrating the first inequality from $0$ to $1$, and the second inequality from $0$ to $+\infty$ we'll get:

$$\int_0^1({1-x^2})^ndx <\int_0^1 e^{-nx^2} dx<\int_0^{\infty} e^{-nx^2} dx<\int_0^{\infty}\frac{dx}{(1+x^2)^n}$$

But,

$$\int_0^{\infty} e^{-nx^2}dx=\frac{1}{\sqrt{n}}K \ \ \ \ \ \ (\text{substitution} \ \ u=\sqrt{nx}),$$

$$\int_0^1({1-x^2})^ndx=\int_0^{\frac{\pi}{2}}\sin^{2n+1}(v)dv=\frac{(2n)!!}{(2n+1)!!} \ \ \ (\text{substitution} \ \ x=\cos(v))$$

and, finally,

$$\int_0^{\infty}\frac{dx}{(1+x^2)^n}=\int_0^{\frac{\pi}{2}}\sin^{2n-2}(v)dv=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2} \ \ \ (\text{substitution} \ \ x=\text{ctg}(v))$$

Hence, our unknown, $K$ is bound:

$$\sqrt{n}\frac{(2n)!!}{(2n+1)!!}<K<\sqrt{n}\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$

or,

$$\frac{n}{2n+1}\frac{((2n)!!)^2}{((2n-1)!!)^2(2n+1)}<K^2<\frac{n}{2n-1}\frac{((2n-3)!!)^2(2n-1)}{((2n-2)!!)^2}(\frac{\pi}{2})^2$$

Now, the final step - Wallis Formula :

$$\lim_{n\to\infty}\frac{((2n)!!)^2}{((2n-1)!!)^2(2n+1)}=\frac{\pi}{2}$$

Then, when $n$ tends to $\infty$ in our last inequality, we get:

$$K^2=\frac{\pi}{4}$$

and,

$$K=\frac{\sqrt{\pi}}{2} \ \ \ \ \ \text{as}\ K>0 $$

Answer 7

By using Beta and Gamma functions properties we may simply obtain that: $$\operatorname B\left(\tfrac 12,\tfrac12\right)=\frac{\left[\Gamma(\tfrac{1}{2})\right]^{2}}{\Gamma{(1)}}=\left[\Gamma(\tfrac{1}{2})\right]^{2}$$ $$\operatorname B\left(\tfrac{1}{2},\tfrac{1}{2}\right)=\frac{\pi}{\sin{\frac{\pi}{2}}}=\pi$$ In other words we have that: $$\Gamma(\tfrac{1}{2})=\sqrt{\pi}\longrightarrow \space\int\limits_0^\infty x^{\frac{-1}{2}} e^{-x} \,\mathrm dx = \sqrt{\pi}$$

By substitution $x=t^2$ we get the final result:

$$2\int\limits_0^\infty e^{-t^2} \,\mathrm dt=\sqrt{\pi} \longrightarrow \int\limits_0^\infty e^{-t^2} \,\mathrm dt = \frac{\sqrt{\pi}}{2}.$$

Q.E.D. (Chris)

Answer 8

Change variables. Let $z=x^2$. We find $\int_{0}^{\infty} e^{-x^2} dx = \frac{1}{2} \Gamma(\frac{1}{2}) = \frac{\sqrt{\pi}}{2}$.

Addendum: Setting $z=1/2$ in Euler's reflection formula, $\Gamma(1-z)\Gamma(z) = \pi/\sin \pi z$, we find $\Gamma(1/2) = \sqrt{\pi}$.

Answer 9

Here you have a solution using Wallis' formula for $\pi$ and the squeeze theorem which I learned. It's from a post I made in a forum.

http://www.mymathforum.com/viewtopic.php?f=15&t=27064

Answer 10

Another way is to make use of the Poisson summation formula. I will work with the Fourier transform $$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) \exp(-2 \pi i \xi x) dx$$ The Poisson summation formula states that $$\sum_{\xi \in \mathbb{Z}} \hat{f}(\xi) = \sum_{n \in \mathbb{Z}} f(n).$$ Now take $f(x) = \exp(-\pi x^2)$. We then get that \begin{align} \hat{f}(\xi) & = \int_{-\infty}^{\infty} \exp(- \pi x^2) \exp(-2 \pi i \xi x) dx\\ & = \int_{-\infty}^{\infty} \exp(-\pi(x+i \xi)^2 - \pi \xi^2) dx\\ & = \exp( - \pi \xi^2)\int_{-\infty}^{\infty} \exp(-\pi(x+i \xi)^2) dx\\ & = \exp( - \pi \xi^2)\int_{-\infty + ic}^{\infty+ic} \exp(-\pi x^2) dx \end{align} By integrating from $-\infty+ic$ to $\infty + ic$, I mean integrate along the line Im$(x) = c$ from left to right. Now since the integrand is analytic, we can move this contour to $X$ axis and conclude that $$\int_{-\infty + ic}^{\infty+ic} \exp(-\pi x^2) dx = \int_{-\infty}^{\infty} \exp(-\pi x^2) dx$$ Hence, we get that $$\hat{f}(\xi) = C \exp( - \pi \xi^2)$$where $C = \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2) dx$. Now make use of the Poisson summation formula to get that $$C \left(\sum_{\xi \in \mathbb{Z}} \exp(-\pi \xi^2) \right) = \left(\sum_{x \in \mathbb{Z}} \exp(-\pi x^2) \right)$$ We can afford to cancel $\displaystyle \left(\sum_{x \in \mathbb{Z}} \exp(-\pi x^2) \right)$ since it converges and hence we can conclude that $$C = \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2) dx = 1$$ Suitable scaling gives you the integral and answer you are looking for.

Answer 11

Using Normal density:

$$ \int_0^\infty e^{-x^2}dx=\sqrt \pi \int_0^\infty \frac{1}{\sqrt \pi}e^{-x^2}dx=\sqrt \pi P(X\geq0), $$

where $X\sim \mathrm{Normal}(0,\frac{1}{2})$. Remember that the normal distribution is symmetric around its mean, so $P(X>0)=P(X>E(X))=\frac{1}{2}$, and the result follows.

Answer 12

Consider the mapping $\eta\! : \mathbb{R}^2\to \mathbb{R}$ given by $$ \eta((x,y)) = \sqrt{x^2+y^2},\quad (x,y)\in\mathbb{R}^2. $$ (1) Show that the image-measure $\lambda_2\circ\eta^{-1}$ is the measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with density $$ f(z)=2\pi z 1_{(0,\infty)}(z),\quad z\in\mathbb{R}, $$ by using Dynkin's Lemma.

(2) Show that $$ \int_{\mathbb{R}^2} e^{-x^2-y^2}\,\lambda_2(\mathrm{d}x,\mathrm{d}y)=\pi $$ by using the formula for integration under measurable transformations.

(3) Use Tonelli's theorem to conclude that $$ \int_{\mathbb{R}}e^{-x^2}\,\lambda(\mathrm{d}x)=\sqrt{\pi}, $$ and now your result follows.

Answer 13

The following argument, similar to Bryan Yock's, is a Feynman parameter trick I invented in Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick

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Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the thing we want to evaluate.

Now note that rather than differentiating directly, it's convenient to multiply by some stuff first to save ourselves some trouble. Specifically, note

$$\left(\frac 1 b e^{-b^2}I\right)' = -2b \int_0^\infty e^{-b^2(1+y^2)} \mathrm d y = -2 e^{-b^2} I(\infty)$$

Then usually at this point we would solve the differential equation for all $b$, and use the known information at the origin to infer the information at infinity. Not so easy here because the indefinite integral of $e^{-x^2}$ isn't known. But we don't actually need the solution in between; we only need to relate information at the origin and infinity. Therefore, we can connect these points by simply integrating the equation definitely; applying $\int_0^\infty \mathrm d b$ we obtain

$$-I'(0)= -2 I(\infty)^2 \quad \implies \quad I(\infty) = \frac{\sqrt \pi} 2$$