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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?

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    $\begingroup$ It might be more correct to say that Solution 2 does not require the Jacobian. You are correct that the equality of the two iterated integrals does require some sort of proof that is beyond single-variable analysis (in this case, the integrand is nonnegative so it's technically Tonelli's theorem that is used). $\endgroup$ – Chappers Jun 29 '18 at 12:33
  • $\begingroup$ @Chappers Thanks for the comment. I tried looking at the wikipedia article, but as I haven't seen any measure theory yet, I am having trouble understanding it $\endgroup$ – DJSmerald Jun 29 '18 at 12:36
  • $\begingroup$ For the purposes of most integrals of explicit functions such as we have here, this is a sufficient specialisation: let $f(x,y)$ be continuous and positive on $(a,b)\times(c,d)$. Then the three integrals $$ \int_{(a,b) \times (c,d)} f(x,y) \,dx\times dy, \quad \int_a^b \left(\int_c^d f(x,y) \, dy\right) dx, \quad\int_c^d\left(\int_a^b f(x,y) \, dx\right) dy $$ are all equal. An advantage of using the Lebesgue integral is that this continues to hold if $a,b,c$ or $d$ is infinite, or one of the integrals is infinite, without extra proof being required, which is not true of the Riemann integral. $\endgroup$ – Chappers Jun 29 '18 at 13:53
  • $\begingroup$ @Chappers Ah I see, thank you so much. So it is very similar to the situation for improper integrals, i.e. we just take limits (e.g. as $a\to -\infty$ and $b\to+\infty$). And Lebesgue integrals allow us to do this without much trouble? $\endgroup$ – DJSmerald Jun 29 '18 at 14:00
  • $\begingroup$ With the Lebesgue integral, one can define an integral on $(-\infty,\infty)$ without taking an extra limit, whereas with the Riemann integral, such an integral has to be defined via a limit. This causes more complication in the proof: you have to not only swap these limits, but swap them with one of the integrals as well (i.e. swap the order of the brackets in $(\lim \int)( \lim \int)$). I suspect the proof of this being possible for reasonably general functions is difficult enough to make it easier to develop a Lebesgue-like integral from scratch and prove Tonelli's theorem. $\endgroup$ – Chappers Jun 29 '18 at 16:29
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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 $(0,\infty) \times (0,\infty)$.

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