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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?

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Check here: – Bombyx mori Aug 29 '12 at 6:55
See also $\int_{\infty}^\infty e^{-x^2}\,dx$ with complex analysis and Proving $\int_{0}^\infty e^{-x^2}\,dx = \frac{\sqrt \pi}{2}$ for a few further methods (most of them without using polar coordinates). – t.b. Aug 29 '12 at 8:37
up vote 11 down vote accepted

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.)

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Method I is really the polar method.. – nbubis Aug 29 '12 at 9:15

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.

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$$ \int_{-\infty} ^\infty e^{-x^2} \; dx =I$$
$$ \int_{-\infty}^\infty e^{-y^2} \; dx =I$$
$$\int_{-\infty}^\infty e^{-x^2}\times e^{-y^2} \; dx =I^2$$
$$\int_{-\infty}^\infty e^{-(x^2+y^2)}\; dx =I^2$$
Polar Coordinates: $x^2+y^2=R^2,-\pi\leqslant \theta\leqslant +\pi$
$$\int_{-\pi}^{+\pi} \int_{0}^ \infty e^{-(R^2)}\; R\times dR \times d(\theta) =I^2$$
$$2\times \pi \int_{0}^ \infty e^{-(R^2)}\; R\times dR =I^2$$
$$-R^2=P\implies -2Rd(R)=d(P)\implies Rd(R)=-d(P)/2$$
$$2\times \pi \int_{-\infty}^ 0 e^{(P)}d(P)/2$$
$$\pi=I^2\Longrightarrow I=\sqrt{\pi}$$
since $e^{x^2}$ is even $\implies$ ans=${\sqrt{\pi}/2}$

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i was looking for other method than changing into polar coordinate. possibly using infinite series – Santosh Linkha Aug 29 '12 at 7:04
still thank you for your effort!! – Santosh Linkha Aug 29 '12 at 7:17

We use the change of variables $y=x^2$. $$ \int_0^\infty e^{-x^2} \; dx = \frac{1}{2}\int_{0}^{\infty} y^{-1/2} {\rm e}^{-y} \,dy = \frac{1}{2} \Gamma(\frac{1}{2}) \,,$$

where $\Gamma(x)$ is defined by the integral, $$\Gamma( x ) = \int_{0}^{\infty} y^{x-1} {\rm e}^{-y} \,dy \,. $$

Series Method

If you are interested in series method, then you can use Watson's lemma;

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I am suspecting the value of Gamma(1/2) is determined according to normal curve – Santosh Linkha Aug 29 '12 at 7:27
maybe you should add $\Gamma(1-z) \; \Gamma(z) = {\pi \over \sin{(\pi z)}} \, $. With $z=1/2$ you get $\Gamma^2(1/2)=\pi$... – draks ... Aug 29 '12 at 7:31
But this does not provide the value of the integral, unless one knows the value of $\Gamma(1/2)$--which is taking the answer for granted, only in a very thinly disguised version. – Did Aug 29 '12 at 9:54

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 measureable 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.

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Let $I=\displaystyle \int_0^{+\infty} e^{-t^2}dt$

Let $f(x)=\displaystyle \int_0^{\tfrac{\pi}{4}} e^{-\tfrac{x^2}{\cos^2t}}dt$

(notice : since $f(x)\leq e^{-x^2}$ then $\lim_{x \rightarrow +\infty}f(x)=0$ )

Let $g(x)=\displaystyle \int_0^x e^{-t^2}dt$


For $x>0$

$\displaystyle x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-x^2\tan^2 t}}{\cos^2t}dt=g(x)$ (1)

(perform change of variable $u=x\tan t$ )

$\displaystyle-2e^{-x^2}g(x)=-2e^{-x^2}\int_0^x e^{-t^2}dt=-2e^{-x^2}x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-x^2\tan^2 t}}{\cos^2t}dt$ (using (1))

$\displaystyle-2e^{-x^2}g(x)=-2x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-x^2(1+\tan^2 t)}}{\cos^2t}dt=-2x\int_0^{\tfrac{\pi}{4}} \dfrac{e^{-\tfrac{x^2}{\cos^2t}}}{\cos^2t}dt=f'(x)$ (2)

$(g(x)^2)'=2e^{-x^2}g(x)=-f'(x)$ (using (2))


$g(x)^2+f(x)$ is constant for all $x$

and $g(0)^2+f(0)=\dfrac{\pi}{4}$

so for all $x$


take the limit of the left member when $x$ goes to infinity:


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

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