Fubini's theorem and the Gaussian Integral I have been reading K. Conrad's very useful monograph on the Gaussian Integral (http://www.math.uconn.edu/~kconrad/blurbs/analysis/gaussianintegral.pdf) but I have a couple of questions which I am having trouble fully justifying to myself:
Let $J=\int^\infty_0 e^{-x^2}dx$ where obviously $2J=\int^\infty_{-\infty} e^{-x^2}dx$ is the more traditional Gaussian integral. In Solution 1, Conrad solves $J$ using polar coordinates, while in Solution 2, he uses the substitution $x=yt$ in the double integral
$J^2=\int^\infty_0 e^{-x^2}dx\int^\infty_0 e^{-y^2}dy=\int^\infty_0(\int^\infty_0 e^{-(x^2+y^2)}dx)dy$.
I don't really have a problem with either solution (except for one point which I will outline below), except that he says his Solution 2 uses only single variable calculus, while Solution 1 uses multivariable calculus. Perhaps it is my own ignorance showing, but why does Solution 2 not use multivariable calculus? Is it because $y$ is a dummy variable?
My other issue comes from the implicit use of Fubini's theorem. Above, where I have written $J^2=...$ I presume it is Fubini that allows me to change the order of integration. However, how does this work when we are working with improper integrals? Further, what would I need to check to make sure Fubini holds? It is easy to show $J$ exists (as a limit), and $e^{-x^2}$ is everywhere non-negative, but is that enough?
 A: In this case you don't need Fubini's theorem or equivalent to produce an iterated integral (although a careless exposition can conceal this fact), since what one is actually doing is using the linearity of the integral twice: you have
$$ J^2 = J\int_0^{\infty} e^{-x^2} \, dx = \int_0^{\infty} e^{-x^2} J \, dx. $$
Now, $e^{-x^2}$ is also a constant when integrating with respect to $y$, so we have
$$ e^{-x^2} J = e^{-x^2} \int_0^{\infty} e^{-y^2}\, dy = \int_0^{\infty} e^{-x^2}e^{-y^2} \, dy, $$
and so
$$ J^2 = \int_0^{\infty} \left( \int_0^{\infty} e^{-(x^2+y^2)} dy \right) dx $$
as desired.
The inner integral $\int_0^{\infty} e^{-(x^2+y^2)} dy$ is now a function of $x$, and we substitute to write this function in a different way, as $x\int_0^{\infty} e^{-x^2(1+t^2)} \, dt $. Now is when you are forced to change the order of integration using some theorem or other. Either we embrace a Lebesgue integral version, such as Tonelli's theorem,

Let $f : (a,b) \times (c,d) \to \mathbb{R}$ be positive and Lebesgue-integrable, where $-\infty \leq a,b \leq \infty$ and $ -\infty \leq c,d \leq \infty$. Then
  $$ \int_{(a,b)\times (c,d)} f(x,y) \, dx \times dy = \int_a^b \left( \int_c^d f(x,y) \, dy \right) dx = \int_c^d \left( \int_a^b f(x,y) \, dx \right) dy. $$

Here, though, we can manage with a simple version for Riemann integrals:

Let $f : (a,b) \times (c,d) \to \mathbb{R}$ be positive and Riemann-integrable, where $-\infty <a,b < \infty$ and $ -\infty < c,d < \infty$. Then
  $$ \int_{(a,b)\times (c,d)} f(x,y) \, dx \times dy = \int_a^b \left( \int_c^d f(x,y) \, dy \right) dx = \int_c^d \left( \int_a^b f(x,y) \, dx \right) dy. $$

In particular, $x = 1-(1-u)^{-1}$ provides a bijection between $(0,1)$ and $(0,\infty)$, so we find that
$$ J^2 = \int_0^1 \int_0^1 \frac{u}{(1-u)^3(1-v)^2}\exp{\left( - \frac{u^2}{(1-u)^2}\left( 1 + \frac{v^2}{(1-v)^2} \right) \right)} \, dv \, du. $$
It turns out that the integrand is in fact continuous on the whole open square, so it's integrable, and we can use the lightweight Tonelli to swap the integrals, and then after undoing the substitutions, we're finally at
$$ J^2 = \int_0^{\infty} \int_0^{\infty} xe^{-x^2(1+t^2)} \, dx \, dt, $$
which is straightforward to do. Alternatively if one is happy to use the Lebesgue integral, we don't need to muck about with substitutions, and can apply the proper version of Tonelli to the integral on $(0,\infty) \times (0,\infty)$.
