Integrate with multiple variables in denominator?

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Homework Help Overview

The discussion revolves around evaluating a double integral with multiple variables in the denominator, specifically the integral \(\int_{0}^{1-v} du \int_{0}^{\frac{1}{2}} dv \frac{1}{1+u^2-v^2}\). Participants are exploring the setup of the integral and potential substitutions or transformations to simplify the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest using a change of coordinates to simplify the integration process, while others discuss the implications of the Jacobian in this context. There are attempts to clarify the original problem's setup and the integration limits, with questions about the correctness of the boundaries in the transformed coordinates.

Discussion Status

The discussion is active, with participants offering various approaches and questioning the assumptions made about the integral's boundaries. Some guidance has been provided regarding coordinate transformations and the need to calculate the Jacobian, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants note discrepancies between the integration ranges in the original and transformed coordinates, indicating potential misunderstandings or errors in the setup. There is also mention of the complexity introduced by the integrand and the transformations being considered.

jketts
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Hey everyone, I need to do the following integral. I just need a little help getting this started, I'm not sure where I need to go. Here is the problem:

[itex]\int_{0}^{1-v} du \int_{0}^{\frac{1}{2}} dv \frac{1}{1+u^2-v^2}[/itex]

I think I have the boundaries for the integral set up correctly, {0≤v≤1/2, 0≤u≤1-v}.
I know that I will have to use [itex]\int dx \frac{1}{c^2+x^2} = \frac{1}{c}tan^ {-1}\frac{x}{2}[/itex]

I began by trying to substitute 1-v^2 as a variable, but then I had to try to integrate tan^-1 with a bunch of square roots in it and that got really bad looking pretty quickly. Thoughts?
 
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I would try a change of coordinates, like x = u+v, y = u-v. The region gets a little more complicated but it's not too bad. Don't forget the Jacobian.
 
The original problem wants to change it from x and y to terms of u and v. I.e., x=u-v, y=u+v.
 
jketts said:
The original problem wants to change it from x and y to terms of u and v. I.e., x=u-v, y=u+v.

Could you give us the original question with the original region?
 
Sure, [itex]\int\int_{D}{} \frac{1}{1+xy} dxdy[/itex] D={(x,y);0≤x≤1,0≤y≤1}, x=u-v, y=u+v. That's all I am given.
 
jketts said:
Sure, [itex]\int\int_{D}{} \frac{1}{1+xy} dxdy[/itex] D={(x,y);0≤x≤1,0≤y≤1}, x=u-v, y=u+v. That's all I am given.

Follow what harup said and at least calculate your jacobian.

##dxdy = \frac{∂(x, y)}{∂(u, v)}dudv##
 
Ok, then you need to look at what you can substitute for v after the first integration. You have ##\frac1{\sqrt{1-v^2}} atan(\sqrt\frac{1-v}{{1+v}})##, right? Let ##\theta = atan(\sqrt\frac{1-v}{{1+v}})##, so ##tan^2(\theta) =\frac{1-v}{1+v}##. When you see tan-squared, what do you think of?
 
jketts said:
Sure, [itex]\int\int_{D}{} \frac{1}{1+xy} dxdy[/itex] D={(x,y);0≤x≤1,0≤y≤1}, x=u-v, y=u+v. That's all I am given.
That integration range doesn't match what you have in u, v coordinates in the OP.
 
[itex]\int_{v}^{1-v}[/itex]

I think I had that part wrong.
 
  • #10
jketts said:
[itex]\int_{v}^{1-v}[/itex]

I think I had that part wrong.

It gets a bit messy. Something like v from -1/2 to +1/2, u from |v| to 1-|v|.
Did you figure out what to do with the tan-squared?
 

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