# When could the product of two integral becomes a double integral?

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?

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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

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:

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

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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