# Square of the Error Function

The author defines the probability integral as follows $$\Phi(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}\mathrm{d}t$$. (S)he instructs the reader to derive the following integral representation of the square of the probability integral by transforming it to polar coordinates $$\Phi(z)^2=1-\frac{4}{\pi}\int_0^1\frac{\exp(-z^2(1+t^2))}{1+t^2}\mathrm{d}t$$. I am uncertain how transforming the integral from cartesian coordinates to polar coordinates would yield the integral representation. Can someone explain this to me? $$I^2=\frac{4}{\pi}\int_{0}^{z}\int_{0}^{z}e^{-(x^2+y^2)}dydx=-\frac{2}{\pi}\int_{0}^\frac{\pi}{2}\int_{0}^{z}-2re^{-r^2}drd\theta$$?

• Solved: en.wikipedia.org/wiki/Gaussian_integral (By the way, I suggest you look up "extricate," as it does not mean what you think it does.) Dec 29 '20 at 0:07
• @DavidG.Stork "Extricate" might have made sense if the previous sentence were something like, "I've gotten stuck in an infinite loop of second-guessing my guesses while trying to figure this out." Dec 29 '20 at 0:16
• Yep! It is one of those words my students use to have that "literate look," but are completely inappropriate. Dec 29 '20 at 0:43
• @DavidG.Stork I know that $$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.$$. However I do not understand how the author derived that $$erf^2(z)=1-\frac{4}{\pi}\int_{0}^{1}\frac{e^{-z^2(1+t^2)}}{1+t^2}dt$$ Dec 29 '20 at 1:33

$$\operatorname{erf}(z)^2$$ can be expressed as an integral over the region $$[0,z]\times[0,z]$$. By symmetry, you can calculate half of its value by looking at the lower half triangle (i.e. with the additional bound $$y\leq x$$). I believe the correct coordinate transformation is $$R=x^2+y^2, t=\frac yx$$ (you could think of the former as $$r^2$$ and the latter as $$\tan\theta$$ in polar coordinates). Express the region in the new coordinates, calculate the Jacobian, and the result should follow.
Full solution: $$\def\pdv#1#2{\frac{\partial #1}{\partial #2}}$$
First, $$erf^2(z)=\frac4\pi\int_{[0,z]^2}e^{-(x^2+y^2)}dxdy=\frac8\pi\int_{D}e^{-(x^2+y^2)}dxdy$$, where $$D=\{(x,y)\in[0,z]^2 \mid y\leq x\}$$. Write $$I=\int_{D}e^{-(x^2+y^2)}dxdy$$.
The inverse of the Jacobian (since this is easier to calculate) is $$\begin{pmatrix}\pdv Rx&\pdv Ry\\ \pdv tx&\pdv ty\end{pmatrix}$$ with determinant $$2(1+t^2)$$, so its inverse has determinant $$\frac1{2(1+t^2)}$$. The triangular region $$D$$ becomes the region $$\{(R,t)|0\leq t\leq1, 0\leq R\leq z^2(t^2+1)\}$$. (You need to notice $$\frac1{\cos^2\theta}=\tan^2\theta+1$$.)
At last the calculation. $$I=\int_0^1\int_0^{z^2(t^2+1)}e^{-R}\cdot\frac1{2(1+t^2)}dRdt=\frac12\int_0^1(1-e^{-z^2(1+t^2)})\frac1{1+t^2}dt=\frac12(\frac\pi4-\int_0^1\frac{e^{-z^2(1+t^2)}}{1+t^2}dt),$$ so $$erf^2(z)=\frac8\pi I=1-\frac4\pi\int_0^1\frac{e^{-z^2(1+t^2)}}{1+t^2}dt$$ as desired.