Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

question1: Why the product of two integral becomes a double integral?

What conditions?

$$\begin{align*}\left(\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty } e^{\frac{-x^2}{2}} \, dx\right)^2=\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty}^{\infty }e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy\\&=\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{\frac{\left(x^2+y^2\right)}{2}}dxdy\end{align*}$$

The integration should be $1$, however when I changed to Polar form, something wrong.

$$\begin{align*}\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy=\frac{1}{2\pi }\int _0^{2\pi }\int_0^{\infty }\color{red}{-}e^{\frac{-r^2}{2}}rdrd\theta \\&=\frac{1}{2\pi }\int _0^{2\pi }e^{\frac{-r^2}{2}}|_0^{\infty }(=-1)d\theta =-1\end{align*}$$

question2:Where goes wrong?

The red Minus $\color{red}{}-\text{}$ is wrong?

question3: Can you show me how to convert $\text{dxdy}$ to $\text{rdrd$\theta $}$ in terms of Jacobians?

share|cite|improve this question
Yes, that's it. $dx\,dy = r\,dr\,d\theta$. – Daniel Fischer Jul 22 '13 at 12:25
Notice in the last equality on the first line, you drop the negative. – Owen Sizemore Jul 22 '13 at 12:26
Where did you take that red minus sign in the middle from? It is wrong... – DonAntonio Jul 22 '13 at 12:29
@DonAntonio a book named <all math you missed>. It gives the answer $1$, and I give $-1$ – User19912312 Jul 22 '13 at 12:31
Well, that's wrong @spuorg-imes...or else you missed another minus sign before the whole double integral... – DonAntonio Jul 22 '13 at 12:35
up vote 1 down vote accepted

This is what you could have missed:

$$\begin{align*}\frac{1}{2\pi }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy=\color{red}-\frac{1}{2\pi }\int _0^{2\pi }\int_0^{\infty }\color{red}{-r}e^{\frac{-r^2}{2}}drd\theta \end{align*}$$

Note that we have then

$$-re^{-\frac{r^2}2}=\frac{d}{dr}\left(-\frac{r^2}2\right)\cdot e^{-\frac{r^2}2}$$

and we use then the following:

$$\int f'(x)e^{f(x)}dx=e^{f(x)}\ldots$$

share|cite|improve this answer
Hi, can you show me how to convert $\text{dxdy}$ to $\text{rdrd$\theta $}$, use Jacobian? and also question1, I'm a newbie – User19912312 Jul 22 '13 at 16:19
@spuorg-imes, as follows: $$x=r\cos\theta\;,\;y=r\sin\theta\implies$$$$ \frac{dx}{dr}=\cos\theta\;,\;\frac{dx}{d\theta}=-r\sin\theta$$$$\frac{dy}{dr}= \sin\theta\;,\;\frac{dy}{d\theta}=r\cos\theta$$ Thus $$\left|J\frac{(x,y)}{(r,\theta)}\right|=\left|\det\begin{pmatrix}\cos\theta&-r \sin\theta\\\sin\theta&r\cos\theta\end{pmatrix}\right|=|r(\cos^2\theta+\sin^2 \theta)|=r$$ – DonAntonio Jul 22 '13 at 19:34
Thanks.~~~~~~~~~~~~~~~~~~ – User19912312 Jul 23 '13 at 0:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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