Changing variables in integrals?

In summary: Then we can express the integrals:∫(v²(x)-vb²)/(va²-vb²) dx = ∫y²(x) dx = ∫(v²(x)-vc²)/(va²-vc²) dxSo the limits of integration for the integral with y² will be the same as the limits for the integral with v².In summary, the authors of the paper make a transition in which they define a new function, y2, to replace the original function, v2, and find a new integral equation for it. Through a series
  • #1
Chen
977
1
Hi,

I'm reading the following paper:
http://prola.aps.org/abstract/PR/v127/i6/p1918_1
(It's physics, you don't really have to click that)

Somewhere along the way the authors make a transition that I can't understand. Basically they have an ODE (5.9), which is integrated to give equation (5.10). Then they define a new function, y2, to replace the original function, v2, and find a new integral equation for it (5.14). I'm bringing two screen shots from the article that show this transition. The only other thing you need to know is that v2a, v2b and v2c are the roots of the expression insider the root in the integral of (5.10).

Screen shots: (in order)
http://img59.imageshack.us/img59/1887/partajj6.png
http://img62.imageshack.us/img62/5359/partbyc9.png

Now, I tried following this transition myself and couldn't make any sense of it. Most of all, I don't understand how the final integral equation (5.14) has y in the limits of integration, and not y2 or something like that.

Help, please? :cry:

Thanks,
Chen
 
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  • #2
Chen, this is straightforward.
Just write v²(1-v²)²-G² = (v²-va²)(v²-vb²)(v²-vc²) = (vb²-va²)y²(v²-va²+va²-vb²),(v²-va²+va²-vc²),
then continue replacing (v²-va²) = y² (vb²-va²) and express everything from y²,
then substitute this is the square root in the denominator,
then don't forget to go from d(v²) to dy (this will do a small simplification,
and its finished.

(unfortunately, I have no access to the original paper)
 
  • #3
lalbatros,

Thanks for the help! You're right, it's pretty straightforward once you know the trick... so I was able to transform the root into the required form, and all the constants also turned out okay - but I'm still bothered by the limits of integration.
Maybe I don't understand something fundamental about this kind of operations, but how do I end up with y(0) and y(xi) in the limits? Starting with v2(0) and v2(xi), can you please explain how these limits transform?
(It's not like I don't know how to change variables inside integrals, but for some reason this seems weird to me...)

Thank you! :-)
Chen
 
  • #4
It is just by chaining the maps!
By definition

y²(x) = (v²(x)-va²)/(vb²-va²)
 

1. What is the purpose of changing variables in integrals?

The purpose of changing variables in integrals is to make the integration process easier or more efficient. By substituting a new variable, the integral may become simpler to evaluate or may lead to a more general solution.

2. How do you choose the appropriate variable to substitute in an integral?

The appropriate variable to substitute in an integral is typically chosen based on the form of the integrand. It should be a variable that simplifies the integrand or leads to a known integral. It may also be chosen based on symmetries or patterns in the integrand.

3. Can changing variables in integrals change the limits of integration?

Yes, changing variables in integrals can change the limits of integration. When substituting a new variable, the limits of integration must also be changed to match the new variable. This is often done by solving for the new limits using the substitution equation.

4. Are there any restrictions on the variables that can be substituted in an integral?

Yes, there may be restrictions on the variables that can be substituted in an integral. The chosen variable must be a continuous and differentiable function of the original variable, and the substitution must result in a one-to-one mapping of the original limits of integration.

5. Can changing variables in integrals be used to solve all integrals?

No, changing variables in integrals cannot be used to solve all integrals. It is a useful technique, but there are some integrals that cannot be solved by any method. Additionally, some integrals may require multiple variable substitutions or other techniques in addition to changing variables.

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