Changing Variables and the Limits of Integration using the Jacobian

In summary, the conversation discusses calculating the Jacobian for converting from x and y variables to u and v variables. The individual is having trouble understanding the limits of integration and how to change them to u and v variables. They attempt to solve the problem by finding the Jacobian and converting the function, but their final answer is incorrect due to not understanding the limits of integration. The conversation ends with a discussion on how to visualize the u and v axes and understanding how they correspond to the first quadrant in the x and y plane.
  • #1
agnimusayoti
240
23
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
agnimusayoti said:
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
$$u = x^2 - y^2$$
$$v=2xy$$
 
  • #4
That one is from the Book
 
  • #5
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
Yeah, that's right. But, wait why that z function (complex function) is necessary? Thanks.
 
  • #7
This is what I do:
I find:
$$f(u,v) = \frac{1}{4} \frac{e^{-v}}{1+u}$$
 
  • #8
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
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
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
Yes. u should run from -∞ to +∞. Do you see why?
 
  • #12
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
Try plotting the curves ##u=\text{constant}## and ##v=\text{constant}## in the ##xy##-plane for various constants.
 
  • #14
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
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
agnimusayoti said:
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
phyzguy said:
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
LCKurtz said:
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
But, I can't see that the ray is transform to ##2\theta##
 
  • #20
agnimusayoti said:
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
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
Thanks a lot member PF.
 
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Related to Changing Variables and the Limits of Integration using the Jacobian

1. What is the Jacobian in the context of changing variables and integration?

The Jacobian is a mathematical concept that represents the rate of change of one set of variables with respect to another set of variables. In the context of changing variables and integration, the Jacobian is used to transform the limits of integration and the integrand to a new coordinate system, making it easier to solve complex integrals.

2. How is the Jacobian calculated?

The Jacobian is calculated by taking the determinant of the Jacobian matrix, which is a matrix of partial derivatives of the new variables with respect to the old variables. The Jacobian matrix is often denoted as J, and the Jacobian as |J|.

3. Why is the Jacobian important in integration?

The Jacobian is important in integration because it allows us to change the variables of integration, making it easier to solve complex integrals. It also helps us to understand the relationship between different coordinate systems and how they affect the integration process.

4. Can the Jacobian be negative?

Yes, the Jacobian can be negative. This means that the orientation of the new coordinate system is opposite to that of the original coordinate system. It is important to consider the sign of the Jacobian when using it to change variables and limits of integration.

5. Are there any limitations to using the Jacobian in integration?

Yes, there are some limitations to using the Jacobian in integration. It is not always possible to find a suitable transformation that will make the integral easier to solve. In addition, the Jacobian may not exist or may be undefined at certain points, which can make it difficult to use in integration.

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