Unknown variable

1. Sep 12, 2007

tomcenjerrym

I am confused about how the y variable came from?

I = e$$^{-x^{2}}$$ dx

I$$^{2}$$ = e$$^{-x^{2}}$$ dx e$$^{-y^{2}}$$ dy

The equation of I should be like this:

I = e$$^{-x^{2}}$$ dx
I$$^{2}$$ = (e$$^{-x^{2}}$$)$$^{2}$$ d$$^{2}$$x
I$$^{2}$$ = e$$^{-2x^{2}}$$ d$$^{2}$$x

2. Sep 12, 2007

genneth

You're missing some integral signs, I believe. Remember that things like dx cannot be separated from the integral sign -- and that in the integrand, the variable x is "bound", and is not the same as an x on the outside. So:

$$I = \int_0^{\infty} e^{-x^2}\,dx = \int_0^{\infty} e^{-y^2}\,dy$$
$$I^2 = \int_0^{\infty} e^{-x^2}\,dx \int_0^{\infty} e^{-y^2}\,dy$$

One more thing: even if you were to be loose with notation and use what I refer to as "physicist shorthand" and treat dx as a variable of sorts, $$dx^2 \ne d^2x$$

3. Sep 12, 2007

tomcenjerrym

and can you tell me why x = y?

4. Sep 12, 2007

It doesn't, but

$$\int_0^{\infty}e^{-x^2}\,dx= \int_0^{\infty}e^{-y^2}\,dy$$

5. Sep 12, 2007

HallsofIvy

Staff Emeritus
$$\int \int f(x) dx dx$$
would make no sense- once you have integrated with respect to x, there is no "x" left to integrate again!

It is true, as DeadWolfe said, that
$$\int_{-\infty}^\infty e^{-x^2}dx= \int_{-\infty}^\infty e^{-y^2}dy[/itex] because the x and y are "dummy" variables. You would integrate with respect to either, then evaluate at the end points (technically, take the limit) so there is no x or y in the final result- it doesn't matter what you call the variable. Of course, it is well known (Fubini's theorem) that the product [tex]\left(\int_a^b f(x)dx\right)\left(\int_c^d g(y)dy$$
is the same as the iterated integral
[tex]\int_{x=a}^b\int_{y=c}^d f(x)g(y) dydx[/itex]