Changing Variables and the Limits of Integration using the Jacobian

  • #1
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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...
 
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  • #2
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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.
 
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  • #3
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$$u = x^2 - y^2$$
$$v=2xy$$
 
  • #4
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That one is from the Book
 
  • #5
LCKurtz
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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?
 
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  • #6
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Yeah, that's right. But, wait why that z function (complex function) is necessary? Thanks.
 
  • #7
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This is what I do:
I find:
$$f(u,v) = \frac{1}{4} \frac{e^{-v}}{1+u}$$
 
  • #8
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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.
 
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  • #9
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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?
 
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  • #10
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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
 
  • #11
phyzguy
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Yes. u should run from -∞ to +∞. Do you see why?
 
  • #12
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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
 
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  • #13
vela
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Try plotting the curves ##u=\text{constant}## and ##v=\text{constant}## in the ##xy##-plane for various constants.
 
  • #14
phyzguy
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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.
 
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  • #15
LCKurtz
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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##.
 
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  • #16
Delta2
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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 $$-r^2\leq r^2\cos 2\theta=u\leq r^2$$
$$-\infty\leq u\leq +\infty$$
 
  • #17
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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?
 

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  • #18
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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.
 
  • #19
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But, I can't see that the ray is transform to ##2\theta##
 
  • #20
phyzguy
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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.
 
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  • #21
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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:
$$I=\frac {1}{4} \int_{v=0}^{\infty} \int_{u=-\infty}^{\infty} \frac {e^{-v}}{1+u^2} du dv = {1/4}(1){\pi}$$
 
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  • #22
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Thanks a lot member PF.
 
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