# improper integral

• Nov 10th 2008, 11:40 PM
iwonder
improper integral
• Nov 10th 2008, 11:51 PM
Moo
Hello,
Quote:

Originally Posted by iwonder

It's a common technique :)
$I=\int_{-\infty}^\infty e^{-3x^2} ~ dx$

Hence $I^2=\left(\int_{-\infty}^\infty e^{-3x^2} ~ dx\right) \cdot \left(\int_{-\infty}^\infty e^{-3y^2} ~ dy\right)$

$I^2=\int_{y=-\infty}^\infty \int_{x=-\infty}^\infty e^{-3x^2} \cdot e^{-3y^2} ~ dx ~ dy$

$I^2=\iint_{\mathbb{R}^2} e^{-3(x^2+y^2)} ~ dx ~ dy$

And you're back to a previous integral
• Nov 10th 2008, 11:56 PM
iwonder
http://www.mathhelpforum.com/math-he...e33ea135-1.gif
and i got the answer for this is pi/3
is not equal to http://www.mathhelpforum.com/math-he...9382d49f-1.gif
plz exaplin(Crying)
• Nov 11th 2008, 12:05 AM
Moo
Quote:

Originally Posted by iwonder
http://www.mathhelpforum.com/math-he...e33ea135-1.gif
and i got the answer for this is pi/3
is not equal to http://www.mathhelpforum.com/math-he...9382d49f-1.gif
plz exaplin(Crying)

It's $I^2$, so take the square root !

$I^2= \frac \pi 3 \implies I= \sqrt{\frac \pi 3}$ (and not -, since the integrand is always positive, its integral is positive)
• Nov 11th 2008, 12:05 AM
bkarpuz
Quote:

Originally Posted by iwonder

If you know the $\Gamma$ function the solution is very simple.

Noting that the function is even, we have
$\int\limits_{-\infty}^{\infty}\mathrm{e}^{-3x^{2}}dx\stackrel{y\to\sqrt{3}x}{=}\frac{2\sqrt{3 }}{3}\int\limits_{0}^{\infty}\mathrm{e}^{-y^{2}}dy$
.................. $\stackrel{z\to y^{2}}{=}\frac{\sqrt{3}}{3}\int\limits_{0}^{\infty }z^{-1/2}\mathrm{e}^{-z}dz$
.................... $=\frac{\sqrt{3}}{3}\Gamma(1/2)$
.................... $=\frac{\sqrt{3\pi}}{3}.$

Also see Gamma function - Wikipedia, the free encyclopedia