how to calculate this integral!!!

**toraj58** · October 23rd 2008, 07:43 AM

how to calulate this :

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

it seems easy but i wonder how it is tricky to solve it.
i planned several approaches but all of them failed. i think the answer should be someting similar to $\pi$ .

**Jhevon** · October 23rd 2008, 07:48 AM

Originally Posted by toraj58

how to calulate this :

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

it seems easy but i wonder how it is tricky to solve it.
i planned several approaches but all of them failed. i think the answer should be someting similar to $\pi$ .

see post #2 here. you should be able to finish up. it shows you how to find $\int_0^\infty e^{-x^2}~dx$

**toraj58** · October 23rd 2008, 09:47 AM

yeah;
that was cool solution.
it was realy nice for me to use double integral trick and then benefit from solving the problem in Polar Coordinates that was undeniably easy.
so the answer is : $\sqrt\pi$

**Jhevon** · October 23rd 2008, 06:46 PM

Originally Posted by toraj58

yeah;
that was cool solution.
it was realy nice for me to use double integral trick and then benefit from solving the problem in Polar Coordinates that was undeniably easy.
so the answer is : $\sqrt\pi$

yes

**Chris L T521** · October 23rd 2008, 06:54 PM

Originally Posted by toraj58

how to calulate this :

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

it seems easy but i wonder how it is tricky to solve it.
i planned several approaches but all of them failed. i think the answer should be someting similar to $\pi$ .

Its also not hard to show that $\Gamma\left(\tfrac{1}{2}\right)=\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}$ , where $\Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$

--Chris

**Moo** · October 23rd 2008, 10:28 PM

Originally Posted by Chris L T521

Its also not hard to show that $\Gamma\left(\tfrac{1}{2}\right)=\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}$ , where $\Gamma(x)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$

--Chris

And how do you know that $\Gamma(\tfrac 12)=\sqrt{\pi}$ ?

_____________________________________
Here is a few other ways.

1.
Consider $g(x)=\int_0^1 \frac{e^{-(t^2+1)x^2}}{t^2+1} ~dt$

$\bullet$ Find $g'(x)$

Let $f(x)=\int_0^x e^{-u^2} ~du$
$\bullet$ Show that $g'(x)=-2f'(x) f(x)$

$\bullet$ Integrate this latter formula.

$\bullet$ Find an expression for $g(x)$ with respect to f(x).

$\bullet$ Show that $\lim_{x \to \infty} g(x)=0$

$\bullet$ Conclude : $\lim_{x \to +\infty} f(x)=\sqrt{\frac{\pi}{4}}$

Since the integrand is even, $\int_{-\infty}^\infty e^{-t^2} ~dt=2 \int_0^\infty e^{-t^2} ~dt$

---------------------------------------
2.
$\bullet$ Show that $\forall x \in ]-1,+\infty[,~ \ln(1+x) \leq x$

$\bullet$ Deduce that $\left(1-\frac{t^2}{n}\right)^n \leq e^{-t^2} \leq \left(1+\frac{t^2}{n}\right)^n$

$\bullet$ Integrate the previous inequality and make substitutions to get, $\forall n \in \mathbb{N}^*$ :
$\sqrt{n} \int_0^{\pi/2} \sin^{2n+1}(u) ~du \leq \int_0^{\sqrt{n}} e^{-t^2} ~dt \leq \sqrt{n} \int_0^{\pi /2} \sin^{2n-2}(u) ~du$

$\bullet$ Admit the Wallis integral : $\int_0^{\pi/2} \sin^p(u) ~du \to \sqrt{\frac{\pi}{2p}} \quad \text{ when } p \to \infty$
And find the value of $\int_0^\infty e^{-t^2} ~dt$

**toraj58** · October 24th 2008, 12:24 PM

thanks for alternative solutions.

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