# Changing variables in integrals?

1. Aug 1, 2007

### Chen

Hi,

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 [Broken]
http://img62.imageshack.us/img62/5359/partbyc9.png [Broken]

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.

Thanks,
Chen

Last edited by a moderator: May 3, 2017
2. Aug 1, 2007

### lalbatros

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.

3. Aug 1, 2007

### Chen

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. Aug 1, 2007

### lalbatros

It is just by chaining the maps!
By definition

y²(x) = (v²(x)-va²)/(vb²-va²)