Changing Variables and the Limits of Integration using the Jacobian

[agnimusayoti]

Homework Statement

Evaluate integral
$$I=\int_{0}^{\infty} \int_{0}^{\infty} \frac {x^2+y^2}{1+(x^2-y^2)^2} e^{-2xy} dx dy$$

Relevant Equations

$J=\frac {\partial {(x,y)}}{\partial {(u,v)}}$

From the equations, I can find Jacobians:
$$J = \frac {1}{4(x^2 + y^2)} $$
But, I confuse with the limit of integration. How can I change it to u,v variables? Thanks...

 

[Mark44]

Homework Statement:: Evaluate integral
$$I=\int_{0}^{\infty} \int_{0}^{\infty} \frac {x^2+y^2}{1+(x^2-y^2)^2} e^{-2xy} dx dy$$
Relevant Equations:: $J=\frac {\partial {(x,y)}}{\partial {(u,v)}}$

From the equations, I can find Jacobians:
$$J = \frac {1}{4(x^2 + y^2)} $$
But, I confuse with the limit of integration. How can I change it to u,v variables? Thanks...

When you calculated the Jacobian, you converted to different variables u and v. What were the conversions you made? Those conversions will determine what the new integration limits will be.

 

[agnimusayoti]

$$u = x^2 - y^2$$
$$v=2xy$$

 

[agnimusayoti]

That one is from the Book

 

[LCKurtz]

I gather you have successfully changed the integral to $u,~v$ variables and are just asking about the limits. Am I correct about that? For doing the limits on this problem, what I have noticed is that $u_x = v_y$ and $v_x = -u_y$ suggesting that $z = x^2 -y^2 + i(2xy)$ is an analytic function, which you may recognize as $f(z) = z^2 = (x + iy)^2$. Looking at your transformation that way, what does the first quadrant transform to under the transformation $z^2$? What are the corresponding $u,~v$ limits?

 

[agnimusayoti]

Yeah, that's right. But, wait why that z function (complex function) is necessary? Thanks.

 

[agnimusayoti]

This is what I do:
I find:
$$f(u,v) = \frac{1}{4} \frac{e^{-v}}{1+u}$$

 

[agnimusayoti]

Then, from
$$ u = x^2 - y^2; v=2xy$$ and for $x,y = 0$ I got $u=0; v=0$.
$$ u = x^2 - y^2; v=2xy$$ and for $x,y = \infty$ I got u as indeterminate form.

 

[agnimusayoti]

I think my problem is I can't draw the new axes from given information. I don't know how u, v axes for this case looks like. Maybe I can start from this? Or?

 

[agnimusayoti]

From relation between first variables (x,y) and newer variable (u,v); I find Jacobians: $$ J = \frac {1}{4 (x^2 + y^2)} =\frac {1}{4 \sqrt{u^2 + v^2}}$$
Then, with these variables, the function becomes:
$$f (u,v) = [\frac {(\sqrt{u^2 + v^2})(e^{-v})}{1+ u^2}] |J| = \frac {e^{-v}}{1+u^2}$$
Then, I try the same limit, which are u and v from 0 to infinite. So, I arrived with the answer $$I = \frac {\pi}{8}$$. But, according to Boas final solution, the answer supposed to be pi/4.

What is my wrong? Was it in the limit of integration? Thankss

 

[phyzguy]

Yes. u should run from -∞ to +∞. Do you see why?

 

[agnimusayoti]

No. I don't see the reason. Because for x and y approach to infinite, i got u an indeterminate form: $u=\infty - \infty$. Please explain, thanksss

 

[vela]

Try plotting the curves $u=\text{constant}$ and $v=\text{constant}$ in the $xy$-plane for various constants.

 

[phyzguy]

The integration volume is the upper right quadrant of the xy plane, where x and y are both positive. So restricting u>0 requires that x>y, which is only half of the integration volume. Plotting the uv curves as @vela suggested should help you see this.

 

[LCKurtz]

Here's another way to see the new limits. Think about $u$ and $v$ in polar coordinates. Then $u = r^2\cos^2\theta - r^2\sin^2\theta = r^2\cos(2\theta)$ and $v = 2r^2\cos\theta \sin\theta = r^2\sin(2\theta)$. Now think about a ray in the first quadrant at angle $\theta$. Hold $\theta$ constant and let $r$ vary. Do you see that that ray transforms to a ray at angle $2\theta$? So if $\theta$ goes from $0$ to $\frac \pi 2$ to cover the first quadrant, $2\theta$ goes from $0$ to $\pi$, which covers the upper half plane. That's why your $u$ variable needs to go from $-\infty$ to $\infty$.

 

[Delta2]

No. I don't see the reason. Because for x and y approach to infinite, i got u an indeterminate form: $u=\infty - \infty$. Please explain, thanksss

In order to get rid of this indeterminate form you got to switch to polar coordinates for x and y, see post #15. Then you got $u=r^2\cos 2\theta$. $\theta$ varies from $0$ to $\pi/2$, so $2\theta$ varies from $0$ to $\pi$, hence $-1\leq\cos 2\theta\leq 1$ and therefore $$-r^2\leq r^2\cos 2\theta=u\leq r^2$$
$$-\infty\leq u\leq +\infty$$

 

[agnimusayoti]

The integration volume is the upper right quadrant of the xy plane, where x and y are both positive. So restricting u>0 requires that x>y, which is only half of the integration volume. Plotting the uv curves as @vela suggested should help you see this.

Here is what I got for the U V axes. Ok, I understand the 1st sentence that the region is in the first quadrant (upper right) of XY plane. But, I can't in u v axes what is equal with first quadrant in XY plane.

However, 2nd explanation implies (for me) that in 1st quadrant, there is 2 option: x > y and x< y but x,y are both positive. To satisfy this, so u must be > 0 and u < 0 (which define the limits : $-\infty<=u<=\infty$. Is that true?

 

[agnimusayoti]

Here's another way to see the new limits. Think about $u$ and $v$ in polar coordinates. Then $u = r^2\cos^2\theta - r^2\sin^2\theta = r^2\cos(2\theta)$ and $v = 2r^2\cos\theta \sin\theta = r^2\sin(2\theta)$. Now think about a ray in the first quadrant at angle $\theta$. Hold $\theta$ constant and let $r$ vary. Do you see that that ray transforms to a ray at angle $2\theta$? So if $\theta$ goes from $0$ to $\frac \pi 2$ to cover the first quadrant, $2\theta$ goes from $0$ to $\pi$, which covers the upper half plane. That's why your $u$ variable needs to go from $-\infty$ to $\infty$.

I started to see that if I use polar coordinate, so XY region is from $\theta_1=0 \rightarrow \theta_2=\frac{\pi}{2}$. Than, if I convert u and v to polar coordinate, the theta now twice from before. So, the region u and v is from 0 to pi. With this in mind, I back to the geometry and see that this region (half XY plane) correspond to u from $u_1 = -\infty \rightarrow u_2 = \infty$ and so does v. Hmm, interesting. So, if the variables used in Jacobian is hyperbolic, I can try polar coordinate to see limit.

 

[agnimusayoti]

But, I can't see that the ray is transform to $2\theta$

 

[phyzguy]

However, 2nd explanation implies (for me) that in 1st quadrant, there is 2 option: x > y and x< y but x,y are both positive. To satisfy this, so u must be > 0 and u < 0 (which define the limits : $-\infty<=u<=\infty$. Is that true?

Yes, I think you understand now.

 

[agnimusayoti]

Well, I think I can answer the question by choosing limit for u from $u_1 = -\infty$ to $u_2 = \infty$ and $v=0$ to $v=\infty$. Limit for v is the same with previous variables because for x,y = 0 I get v = 0 and for x,y approach $\infty$, v approach $\infty$.

According previous post, there are 2 methods for define the limit for u since I get indeterminate form while substitute the value for x and y approach infite. I call it geometrically and analytically. With geometric, I have to draw x,y plane and see the region. Than, in the same plane, I have to draw u,v axes and see that in first quadrant, there is one condition that supposed to be satisfied: x,y both positives. That means, x can bigger than y or x is smaller than y. If u varies from 0 to infinite than this limit only a half of original region. So, u is supposed to vary from $-\infty$ to $\infty$.

Analytically, I have to use polar form to see that first quadrant is corresponded with $\theta$ varies from 0 to $\pi/2$. Then, I have to change u variable to polar form and see that $\theta$ is transformed to $2\theta$ by this axe. So, in this variable, $\theta$ varies from 0 to $2(\pi/2)$. This region is corresponded with circle with radius from $-\infty$ to $\infty$. So, this is the u limit.

For the v limit, I can just take limit $v=2xy$ as x,y approach 0 and infinite. That's it. So, the integral becomes to:
$$I=\frac {1}{4} \int_{v=0}^{\infty} \int_{u=-\infty}^{\infty} \frac {e^{-v}}{1+u^2} du dv = {1/4}(1){\pi}$$

 

[agnimusayoti]

Thanks a lot member PF.