Strange change of variables

  • Thread starter LAHLH
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  • #1
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Hi,

I'm reading an article where an integral of the form:

[tex] \int^{\infty}_{-\infty}\,\mathrm{d}\tau' \int^{\infty}_{-\infty}\,\mathrm{d}\tau''...[/tex]

The author then splits this into the region whereby [tex]\tau' >\tau'' [/tex], and the region [tex] \tau''>\tau' [/tex].

[tex] \int^{\infty}_{-\infty}\,\mathrm{d}\tau' \int^{\infty}_{\tau'}\,\mathrm{d}\tau''...+\int^{\infty}_{-\infty}\,\mathrm{d}\tau' \int^{\tau'}_{-\infty}\,\mathrm{d}\tau''...[/tex]

For the first integral the change of variables [tex] u=\tau'',s=\tau''-\tau' [/tex] is made, (and for the second part the change [tex] u=\tau',s=\tau'-\tau'' [/tex])

Focusing on just say the first integral for clarity, is this change of variables really well defined? I mean we could write [tex] \tau' =u-s, \tau''=u [/tex] and then we arrive at [tex] \mathrm{d}\tau'=du-ds, \mathrm{d}\tau''=du [/tex]. How does one uniquely change the s integral measure? My instincts say I should just use [tex] \mathrm{d}\tau'' \rightarrow du, \mathrm{d}\tau'\rightarrow -ds [/tex] but I don't really know how to justify this.

Secondly the limits, starting with the [tex] \tau'' [/tex] integral, [tex] \tau'' \in [\tau', +\infty]\rightarrow u \in [u-s ,+\infty] [/tex]....but the variable u can't be in it's own lower limit??

I can see by eye, if you like, that the max and min values of [tex] \tau' [/tex] are going to be [tex] u \in [-\infty,+\infty], s \in [0,\infty] [/tex], and indeed this is what the author has written, but how to show this analytically?

I tried to use Maple, just to see if it could do it using the change of variables commands, but it said it was "unable to solve the change of variable equations"
 

Answers and Replies

  • #2
Mute
Homework Helper
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I mean we could write [tex] \tau' =u-s, \tau''=u [/tex] and then we arrive at [tex] \mathrm{d}\tau'=du-ds, \mathrm{d}\tau''=du [/tex]. How does one uniquely change the s integral measure? My instincts say I should just use [tex] \mathrm{d}\tau'' \rightarrow du, \mathrm{d}\tau'\rightarrow -ds [/tex] but I don't really know how to justify this.

When changing variables in a multi-integral, that's not how it works. In the double integral case you sort of have to think of it as an area integral dA, which in the current coordinates [itex]dA = d\tau'd\tau''[/itex]. When you change variables, though, dA needs to be expressed in terms of the new variables, [itex]dA = |\mathcal J| du ds[/itex], where [itex]|\mathcal J(u,s)|[/itex] is something called the Jacobian (or jacobian determinant, to be technically correct), which describes the scaling factor that accounts for the change of variables. You compute the jacobian as:

[tex]|\mathcal J(s,u)| = \left|\begin{array}{c c} \frac{\partial \tau'}{\partial u} & \frac{\tau''}{\partial u} \\ \frac{\partial \tau'}{\partial s} & \frac{\partial \tau''}{\partial s} \end{array}\right|[/tex]

Of course, since your changes of variables are rather simple (one of the variables you're just relabeling), you can get away with doing the one dimensional sort of thing: in the first term, the first integral is

[tex]\int_{\tau'}^\infty d\tau''[/itex]

As far as this integral is concerned, you can treat [itex]\tau'[/itex] as a constant:

[tex]\int_{\tau'-\tau'}^\infty d(\tau''-\tau') = \int_{0}^\infty ds[/itex]

But if the change of variables isn't that simple, you need to use the jacobian method.
 
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  • #3
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Thanks alot for the reply.
 

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