2
$\begingroup$

I am simply confused because of planes now.

Consider:

$$J = \int_{R}\int_{R} e^{-(x^2 + y^2)} dxdy$$

What is the geometrical aspect of this integral?

This represents the volume under $h(x,y) = e^{-(x^2 + y^2)}$ I believe.

Then how is:

$$I = \int_{R} e^{-x^2} dx = \int_{R} e^{-y^2} dy$$?

How does that statement hold true? Which are independent/dependent variables? I am utterly confused.

$\endgroup$
15
  • $\begingroup$ The formula involving $I$ seems to be just a definition of $I$. The second equality is trivially true: It reflects the fact that it doesn't matter what letter you use for the integration variable in an integral. The interesting question is why is $J=I^2$ … $\endgroup$ – Harald Hanche-Olsen Jan 6 '15 at 19:00
  • $\begingroup$ It actually is true. Here:mathforum.org/library/drmath/view/69832.html $\endgroup$ – anonymous Jan 6 '15 at 19:01
  • 1
    $\begingroup$ I'm hoping that someone else will step up and explain it. I am very bad at clearing up confusions at this level, unless I have the confused person and a blackboard at hand. $\endgroup$ – Harald Hanche-Olsen Jan 6 '15 at 19:06
  • 1
    $\begingroup$ Er, no, there is no relation indicated between $x$ and $y$ except that they occur in a similar manner in the formula. $\endgroup$ – Harald Hanche-Olsen Jan 6 '15 at 19:08
  • 1
    $\begingroup$ The surface is $z=e^{-(x^2+y^2)}$. No, it does not touch the $xy$-plane. At all points $(x,y)$ we have $z>0$. $\endgroup$ – Jyrki Lahtonen Jan 6 '15 at 19:13
3
$\begingroup$

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

This is an important integral in probability theory, so it pays to understand what is happening here. You can't get the indefinite integral in terms of elementary functions, but you can find the value of the integral with limits $-\infty$ to $+\infty$.

First, the statement that you are confused about is, as Harald says, merely a matter of switching what you call the integration variable; that does not change the value of the integral. Try it: What is $$ \int_0^1 x\, dx$$ and what is $$\int_0^1 v\, dv$$ You can see the integration variable disappears in the end.

Next, why in the world are we introducing that equivalent but differently "named" integral $\int_{-\infty}^\infty e^{-y^2} dy$? We are doing it just so that we can multiply the two integrals together, getting $$ I^2 = \int_{-\infty}^\infty e^{-x^2} dx \int_{-\infty}^\infty e^{-y^2} dy $$ and then since when doing the $x$ integral the entire $y$ integral is just a constant (it does not depend on $x$, after all), we can re-write this as a double integral $$I^2 = \int_{x=-\infty}^\infty \int_{y=-\infty}^\infty e^{-(x^2+y^2)} dy\,dx $$

Why is this progress? The trick is (and I like to think of this as having been discovered by Gauss, but probably others did this before him) that this double integral is over the whole $x,y$ plane, and we can therefore easily transform to polar coordinates, because the limits will be $r$ going from $0$ to $\infty$ and $\theta$ from $0$ to $2\pi$.

The $(x^2+y^2)$ is conveniently $r^2$, and -- here is the zinger -- $$ dx\,dy -\longrightarrow r\,d\theta\,dr $$

That extra factor of $r$ is golden because now we can do the integral: $$ I^2 = \int_{\theta = 0}^{2\pi} d\theta \int_{r=0}^{\infty} r e^{-r^2} dr = 2\pi \left[ -\frac{1}{2} e^{-r^2} \right]_{r=0}^{r=\infty} = \pi(1-0) = \pi $$ SO $$ I = \sqrt{\pi}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.