# How do I compute this integral with a Dirac's delta?

While studying probability I encountered this integral $$I=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2$$ If I compute this in polar coordinates i get $$I=\int_0^{2\pi}\,d\theta \int_0^{+\infty}\exp\left(-\dfrac{\rho^2}{2} \right)\rho\delta(r-\rho)\,d\rho=2\pi r\exp\left(-\dfrac{r^2}{2}\right)$$ but in cartesian coordinates I only get $$I=\exp\left(-\frac{r^2}{2}\right)$$ I don't understand why. I just thougth that I was using the Dirac's delta's properties in both cases. I think the first result is the correct one and there is something I don't know about Dirac's delta with more than one variable.

Which result is correct and why?

You may also make direct calculations in Cartesian coordinates - integrating, for instance, over $$x_1$$ first and then over $$x_2$$. We can multiply and divide the argument of $$\delta$$-function ($$r-\sqrt{x_1^2+x_2^2}$$) by ($$r+\sqrt{x_1^2+x_2^2}$$) (because ($$r+\sqrt{x_1^2+x_2^2}$$) is always positive). We can also replace the power of exponent by $$-\frac{r^2}{2}$$ - due to the condition imposed by $$\delta$$-function

$$I(r)=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2=\int_{\mathbb{R}^2}\exp\left({-\frac{r^2}{2}}\right)\delta\left(\frac{r^2-(x_1^2+x_2^2)}{r+\sqrt{x_1^2+x_2^2}}\right)\,dx_1\,dx_2$$

We see that $$x_2$$ contributes if only $$x_2\in[-r,r]$$, otherwise $$\delta()=0$$

$$I(r)=\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)\delta\left(\frac{(\sqrt{r^2-x_2^2}-x_1)(\sqrt{r^2-x_2^2}+x_1)}{r+\sqrt{x_1^2+x_2^2}}\right)$$

But $$\delta(\frac{(a-x_1)(x_1+b)}{A})=|\frac{A}{a-x_1}|\delta(x_1+b)+|\frac{A}{x_1+b}|\delta(a-x_1)=|\frac{A}{a-x_1}|\delta(x_1+b)+|\frac{A}{x_1+b}|\delta(x_1-a)$$

We get

$$I(r)=\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)\left(r+\sqrt{x_1^2+x_2^2}\right)\left(\frac{1}{\sqrt{r^2-x_2^2}-x_1}\delta(\sqrt{r^2-x_2^2}+x_1)+\frac{1}{\sqrt{r^2-x_2^2}+x_1}\delta(\sqrt{r^2-x_2^2}-x_1)\right)=$$ $$\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)2r\left(\frac{1}{2\sqrt{r^2-x_2^2}}\delta(\sqrt{r^2-x_2^2}+x_1)+\frac{1}{2\sqrt{r^2-x_2^2}}\delta(x_1-\sqrt{r^2-x_2^2})\right)=\int_{-r}^r\exp\left({-\frac{r^2}{2}}\right)\frac{2r}{\sqrt{r^2-x_2^2}}dx_2$$ $$I(r)=\int_{-1}^1\exp\left({-\frac{r^2}{2}}\right)\frac{2r}{\sqrt{1-t^2}}dt=2\pi{r}e^{-\frac{r^2}{2}}$$

In Cartesian coordinates, you have \begin{align} I=&\ \int dx\ \exp\left(-\frac{x_1^2+x_2^2}{2} \right)\delta(r-\sqrt{x_1^2+x_2^2})\\ =&\ \int_{r=\sqrt{x_1^2+x_2^2}} d\sigma\ \exp\left(-\frac{x_1^2+x_2^2}{2} \right) \end{align} since \begin{align} |\nabla (r- \sqrt{x_1^2+x_2^2})| = 1. \end{align} Hence, it follows that \begin{align} I = \int_{r=\sqrt{x_1^2+x_2^2}} d\sigma\ \exp\left(-\frac{r^2}{2} \right) = 2\pi r\exp\left(-\frac{r^2}{2} \right). \end{align}

• Is it like a surface integral in one dimension instead of two? Why did you compute the gradient? Feb 15, 2021 at 21:53
• @Rhino Please consult en.wikipedia.org/wiki/…. Feb 15, 2021 at 22:00