4
$\begingroup$

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? $$I^2=\frac{4}{\pi}\int_{0}^{z}\int_{0}^{z}e^{-(x^2+y^2)}dydx=-\frac{2}{\pi}\int_{0}^\frac{\pi}{2}\int_{0}^{z}-2re^{-r^2}drd\theta$$?

$\endgroup$
4
  • 2
    $\begingroup$ 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.) $\endgroup$ Dec 29 '20 at 0:07
  • 5
    $\begingroup$ @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." $\endgroup$
    – David K
    Dec 29 '20 at 0:16
  • 2
    $\begingroup$ Yep! It is one of those words my students use to have that "literate look," but are completely inappropriate. $\endgroup$ Dec 29 '20 at 0:43
  • 1
    $\begingroup$ @DavidG.Stork I know that $$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.$$. However I do not understand how the author derived that $$erf^2(z)=1-\frac{4}{\pi}\int_{0}^{1}\frac{e^{-z^2(1+t^2)}}{1+t^2}dt$$ $\endgroup$
    – Poltroon
    Dec 29 '20 at 1:33
5
$\begingroup$

$\operatorname{erf}(z)^2$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.