An integral involving the error function I have in my notes the following problem. I recall it being quite difficult and needing a change of variables into polar or spherical coordinates. Assuming I have not made a typo, there is a nice exact answer waiting at the end. 
I suspect the last integral might actually be to the $-x^2$ power. Does it even converge? I am posting this problem because I think you will find it interesting, and because I am trying to make sure my notes are correct.

Define $$I(x)=\int_x^\infty e^{-y^2} \ dy$$
Evaluate $$\int_0^\infty e^{x^2} [ I(x) ]^2 \ dx$$

 A: I will show below that there is no need to use polars or any other coordinates other than rectangulars to evaluate this integral.  The trick here is to rescale the integral in $I(x)$ so that it is easier to manipulate later.  To wit
$$I(x) = x \int_1^{\infty} du\: e^{-x^2 u^2}$$
which means that
$$I(x)^2 = x^2 \int_1^{\infty} du\: \int_1^{\infty} dv\: e^{-x^2 (u^2+v^2)}$$
Now we can put this into the desired integral and reverse the order of integration to get
$$\begin{align}\int_0^{\infty} dx \: e^{x^2} I(x)^2 = \int_1^{\infty} du\: \int_1^{\infty} dv\: \int_0^{\infty} dx \: x^2 e^{-x^2 (u^2+v^2-1)}\end{align}$$
The inner integral converges because $u^2+v^2-1 \ge 0$.  We evaluate the inner integral and reduce the integral to a double integral:
$$\int_0^{\infty} dx \: e^{x^2} I(x)^2 = \frac{\sqrt{\pi}}{4} \int_1^{\infty} du\: \int_1^{\infty} dv\: (u^2+v^2-1)^{-3/2}$$
The integral over $v$ may be attacked by a trig substitution: $v=\sqrt{u^2-1} \tan{t}$, $dv = \sqrt{u^2-1} \sec^2{t} dt$; the result is
$$\int_1^{\infty} dv\: (u^2+v^2-1)^{-3/2} = \left ( 1 - \frac{1}{u}\right ) \frac{1}{u^2-1} = \frac{1}{u (u+1)}$$
The problem now reduces to the evaluation of 
$$\int_1^{\infty} \frac{du}{u (u+1)} = \lim_{r \rightarrow \infty} \log{\left(\frac{r}{r+1}\right)} + \log{2} = \log{2}$$
Therefore, the desired integral has the value
$$\int_0^{\infty} dx \: e^{x^2} I(x)^2 = \frac{\sqrt{\pi}}{4} \log{2} $$
ADDENDUM
This result may be checked against a source such as Wolfram Alpha by considering
$$\int_0^{\infty} dx \: e^{x^2} \, \text{erfc}^2{x}$$
It turns out that $\text{erfc}{x} = (2\sqrt{\pi}) I(x)$.  The result one would expect from WA, then is $\log{2}/\sqrt{\pi}$, which is what WA produces.
