Change of variables in two dimensions

Question

This is from Munkres' Analysis on Manifolds, Section 17, Question 4.

(a) Show that $$ \int_\Bbb {R^2} e^{-(x^2+y^2)} = \left[ \int_\Bbb R e^{-x^2}\right]^2,$$ provided the first of these integrals exists.

(b) Show that the first of these integrals exists and evaluate it.

So using the change of variable equation with $g(x,y) = (x\cos(y), x\sin(y))$, you get $$\int e^{-x^2}x.$$ However, this does not even equal the same thing as the original integral.

Answer 1

This is known as the problem of the Gaussian Integral. Some very useful information on this problem, including the careful derivation and evaluation, can be found here.

Answer 2

a) By Fubini's theorem, we have, if

$$ \int_\Bbb {R^2} e^{-(x^2+y^2)}dxdy < \infty, $$

then

$$ \int_\Bbb {R^2} e^{-(x^2+y^2)}dxdy = \int_\Bbb {R^2} e^{-(x^2+y^2)}dydx= \int_\Bbb {R} e^{-x^2}dx \int_\Bbb {R}e^{-y^2}dy = \left(\int_\Bbb {R} e^{-x^2}dx \right)^2 .$$

b) Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$, then we have

$$ \int_\Bbb {R^2} e^{-(x^2+y^2)}dxdy = 4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^2+y^2)}=4 \int_{0}^{\pi/2}\int_{0}^{\infty} e^{-r^2} r dr d\theta= \pi .$$

Answer 3

You are trying to find the integral:

$$I := \int_{-\infty}^{\infty} \operatorname{e}^{-x^2} \, \operatorname{d}\!x$$

The trick is to consider the following product:

$$\left( \int_{-\infty}^{\infty} \operatorname{e}^{-x^2} \, \operatorname{d}\!x \right)\left( \int_{-\infty}^{\infty} \operatorname{e}^{-y^2} \, \operatorname{d}\!y \right) = I^2$$

We can bring these two integrals together to form a double integral:

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \operatorname{e}^{-x^2} \operatorname{e}^{-y^2}\operatorname{d}\!x \, \operatorname{d}\!y = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \operatorname{e}^{-(x^2+y^2)} \operatorname{d}\!x \, \operatorname{d}\!y$$

To evaluate the double integral, we make the following change of variable: $x=r\cos\theta$ and $y=r\sin\theta$. In such a case, $x^2+y^2 = r^2$ and

$$\begin{array}{ccc} \operatorname{d}\!x & = & \cos\theta \, \operatorname{d}\!r - r\sin\theta \, \operatorname{d}\!\theta \\ \operatorname{d}\!y & = & \sin\theta \, \operatorname{d}\!r + r\cos\theta \, \operatorname{d}\!\theta \end{array}$$

It follows that $\operatorname{d}\!x \, \operatorname{d}\!y = \operatorname{d}\!x \wedge \operatorname{d}\!y = r \, \operatorname{d}\!r \wedge \operatorname{d}\!\theta $ Hence:

$$I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \operatorname{e}^{-(x^2+y^2)} \operatorname{d}\!x \, \operatorname{d}\!y = \int_0^{2\pi}\int_0^{\infty} r\operatorname{e}^{-r^2} \, \operatorname{d}\!r \, \operatorname{d}\!\theta$$

The change of limits came from the fact that, in polar coordinates, a ray from the origin to positive infinity, rotated about the origin by an angle of $2\pi$ radians will cover the plane. Finally:

\begin{array}{ccc} \int_0^{2\pi}\int_0^{\infty} r\operatorname{e}^{-r^2} \, \operatorname{d}\!r \, \operatorname{d}\!\theta &=& \int_0^{2\pi}\left[-\frac{1}{2}\operatorname{e}^{-r^2}\right]_0^{\infty} \operatorname{d}\!\theta \\ &=& \frac{1}{2}\int_0^{2\pi} \operatorname{d}\!\theta \\ &=& \pi \end{array}

We finally conclude that $I^2 = \pi$ and, since $\operatorname{e}^{-x^2} > 0$ for all $x$, hence $I = \sqrt{\pi}$.

Answer 4

I am reading "Analysis on Manifolds" by James R. Munkres.

I used the following propositions:

Theorem 15.2 on p.123 in "Analysis on Manifolds"
Let $A$ be open in $\mathbb{R}^n$; let $f:A\to\mathbb{R}$ be continuous. Choose a sequence $C_N$ of compact rectifiable subsets of $A$ whose union is $A$ such that $C_N\subset\operatorname{Int}C_{N+1}$ for each $N$. Then $f$ is integrable over $A$ if and only if the sequence $\int_{C_N} |f|$ is bounded. In this case, $$\int_A f=\lim_{N\to\infty} \int_{C_N} f.$$

My Proposition:
If $\lim_{n\to\infty}x_n^2$ exists, then $\lim_{n\to\infty} |x_n|$ exists.

I solved this exercise as follows:

(a)
Let $D_N:=\{(x,y)\in\mathbb{R}^2\mid -N\leq x\leq N\text{ and }-N\leq y\leq N\}.$
Then the union of a sequence $D_N$ of compact rectifiable subsets of $\mathbb{R}^2$ is $\mathbb{R}^2$ such that $D_N\subset\operatorname{Int}D_{N+1}$ for each $N$.
And $\int_{\mathbb{R}^2} e^{-(x^2+y^2)}$ exists by our assumption.
So, by Theorem 15.2, $\lim_{N\to\infty} \int_{D_N} e^{-(x^2+y^2)}$ exists and $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\lim_{N\to\infty} \int_{D_N} e^{-(x^2+y^2)}.$$
$\int_{D_N} e^{-(x^2+y^2)}=\int_{y=-N}^{y=N}\int_{x=-N}^{x=N} e^{-(x^2+y^2)}=\int_{y=-N}^{y=N}e^{-y^2}\int_{x=-N}^{x=N}e^{-x^2}=\left(\int_{x=-N}^{x=N}e^{-x^2}\right)^2.$
So, $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\lim_{N\to\infty}\left(\int_{x=-N}^{x=N}e^{-x^2}\right)^2.$$
By my proposition above, $\lim_{N\to\infty}\int_{x=-N}^{x=N}e^{-x^2}$ exists and $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\left(\lim_{N\to\infty}\int_{x=-N}^{x=N}e^{-x^2}\right)^2.$$
Let $B_N:=\{x\mid -N\leq x\leq N\}.$
Then the union of a sequence $B_N$ of compact rectifiable subsets of $\mathbb{R}$ is $\mathbb{R}$ such that $B_N\subset\operatorname{Int}B_{N+1}$ for each $N$.
So, by Theorem 15.2, $\int_{\mathbb{R}}e^{-x^2}$ exists and $$\int_{\mathbb{R}}e^{-x^2}=\lim_{N\to\infty}\int_{x=-N}^{x=N}e^{-x^2}.$$
So, $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\left[\int_{\mathbb{R}}e^{-x^2}\right]^2.$$

(b)
Let $C_N:=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2\leq N^2\}.$
Then the union of a sequence $C_N$ of compact rectifiable subsets of $\mathbb{R}^2$ is $\mathbb{R}^2$ such that $C_N\subset\operatorname{Int}C_{N+1}$ for each $N$.
Let $U_N:=\{(x,y)\mid x^2+y^2<N^2\text{ and }x<0\text{ if }y=0\}.$
The set $U_N$ consists of $\operatorname{Int}C_N$ with the non-negative $x$-axis deleted. Because the non-negative $x$-axis has measure zero, $$\int_{C_N} e^{-(x^2+y^2)}=\int_{\operatorname{Int}C_N} e^{-(x^2+y^2)}=\int_{U_N} e^{-(x^2+y^2)}.$$
The polar coordinate transformation $g$ defines a diffeomorphism of the open set $(0,N)\times (0,2\pi)$ with $U_N$.
So, by Change of variables theorem, $$\lim_{N\to\infty}\int_{C_N} e^{-(x^2+y^2)}=\lim_{N\to\infty}\int_{U_N} e^{-(x^2+y^2)}=\lim_{N\to\infty}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=N} r e^{-r^2}=\pi.$$
So, by Theorem 15.2, $\int_{\mathbb{R}^2} e^{-(x^2+y^2)}$ exists and $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\lim_{N\to\infty}\int_{C_N} e^{-(x^2+y^2)}=\pi.$$