Exercise [12.10]

DimBulb · 29 Jul 2009, 19:14

Let $G = \int_{-\infty}^{\infty} e^{-x^{2}} dx$ . Explain why $G^{2} = \int_{\Re^{2}} e^{-(x^{2}+y^{2})} dx \wedge dy$ and evaluate this by chaning to polar coordinates $(r,\theta)$ .

Hence prove $G = \sqrt{\pi}$ .
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This, $e^{-x^{2}} dx$ , is a 1-form.

So is this $e^{-y^{2}} dy$ .

If $G = \int_{-\infty}^{\infty} e^{-x^{2}} dx$ , then also $G = \int_{-\infty}^{\infty} e^{-y^{2}} dy$ because the

or

is a dummy variable that disappears with integration.

So multiplying these two ways of expressing

together would give us G^2

$G^2 = \int_{-\infty}^{\infty} e^{-x^{2}} dx \int_{-\infty}^{\infty} e^{-y^{2}} dy$

This part is a little confusing to me. We don't multiply 1-forms, we "wedge product" them. Is that justification enough to go to the next step as follows,

$G^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}} dx \wedge e^{-y^{2}} dy$ ?

$e^{-x^{2}} dx \wedge e^{-y^{2}} dy = e^{-x^{2}} e^{-y^{2}} dx \wedge dy = e^{-(x^{2}+y^{2})} dx \wedge dy$ is a 2-form, appropriate for integrating over a plane, as Penrose tells us (See page 231 and note 12.9 on page 245). A little confusion remains on my part.

$G^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}} e^{-y^{2}} dx \wedge dy$

$G^2 = \int_{\Re^{2}} e^{-(x^{2}+y^{2})} dx \wedge dy$

Changing to polar coordinates:

$r = \sqrt{x^2 +y^2}$
$\theta = \sin^{-1} \left( \frac{y}{\sqrt{x^2+y^2}} \right)$

$x = r \cos \theta$
$y = r \sin \theta$
$dx = \cos \theta{ }dr - r \sin\theta{ } d\theta$
$dy = \sin \theta{ } dr + r \cos\theta{ }d\theta$

$dx\wedge dy =$
$\left(r \cos^2 \theta \right)dr \wedge d\theta - \left(r \sin^2 \theta \right) d\theta \wedge dr$
$= r\left( \cos^2 \theta + \sin^2 \theta \right) dr \wedge d\theta$
$= r dr \wedge d \theta$

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^{2}+y^{2})} dx \wedge dy$

$= \int_{-\infty}^{\infty} \int_{0}^{\infty} e^{-(x^{2}+y^{2})} dx \wedge dy + \int_{-\infty}^{\infty} \int_{-\infty}^{0} e^{-(x^{2}+y^{2})} dx \wedge dy$

$= \int_{-\infty}^{\infty} \int_{0}^{\infty} e^{-(x^{2}+y^{2})} dx \wedge dy + \int_{-\infty}^{\infty} \int_{0}^{\infty} e^{-(x^{2}+y^{2})} dx \wedge dy$

$= \int_{-\infty}^{\infty} \int_{0}^{\infty} 2 e^{-(x^{2}+y^{2})} dx \wedge dy$

$= \int_{\theta (y = -\infty) }^{\theta (y = \infty)} \int_{r(x =0)}^{r(x = \infty)} 2r e^{-r^{2}} dr \wedge d\theta$

$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\infty} 2r e^{-r^{2}} dr \wedge d\theta$

The change in integration limits justified because for $\theta$

$\sin^{-1} \left( \frac {-\infty}{\sqrt{\left(-\infty\right)^2 + x^2}} \right) \rightarrow -\frac{\pi}{2}$ (OK, I mean this is the limit as

approaches $-\infty$ )

$\sin^{-1} \left( \frac { \infty}{\sqrt{ \infty^2 + x^2}} \right) \rightarrow \frac{\pi}{2}$ ( for

going to $\infty$ )

and for

$\sqrt {\left(\infty^2 + y\right)} \rightarrow \infty$ (for x going to $\infty$ )

What about the lower limit of

corresponding to when x=0

? We are integrating along the half line from a point where x=0

to $x=+\infty$ in the

integration. I think integrating along any line to infinity (except along the

-axis in which case

would not change) would do the trick, just so long as the angle $\theta$ remains constant in this integration. If we started at any other point than the origin on the

-axis,

at the start of this integration, but the angle $\theta$ would have to change as

increased towards $\infty$ . That is about as good a reason as I can come up with as to why x=0

for a lower limit means r=0

for a lower limit with the variable change.

Continuing, and substituting $u = -r^2, du = -2rdr, r=\infty \rightarrow u = -\infty$
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\infty} 2r e^{-r^{2}} dr \wedge d\theta$
$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{-\infty} -e^{u} du \wedge d\theta$

$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( -e^{-\infty} - -e^{0} \right) d\theta$ (I am a little uncertain about this step as well. I mean losing the wedge. Is it justified?)

$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 d\theta = \frac{\pi}{2} - -\frac{\pi}{2} = \pi$

So

$G^2 = \pi$ ,
and
$G = \sqrt\pi$