Double Integration change of variables

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Discussion Overview

The discussion revolves around a transformation involving double integration in the context of calculus of variations, specifically concerning the displacement of an optimal path. Participants explore the equivalence of two integral expressions defined over the interval from 0 to T, with a focus on the implications of changing variables within the integrals.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents an integral transformation involving a function \phi(t) and seeks clarification on its validity.
  • Another participant suggests that the transformation is valid due to the interchangeable nature of the variables t and \tau, emphasizing that the only difference in the integrands is the swapping of these variables.
  • The second participant notes the importance of correctly defining the range of integration when changing variables, specifically mentioning the triangular area defined by 0 < \tau < t and 0 < t < T.
  • The same participant concludes that swapping t and \tau results in a new range of integration where t < \tau < T.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the transformation, with one participant expressing satisfaction with the explanation provided. However, the discussion does not explore any competing views or unresolved issues regarding the transformation itself.

Contextual Notes

Participants do not explicitly address any limitations or assumptions regarding the transformation or the integrals involved.

Who May Find This Useful

This discussion may be useful for individuals interested in calculus of variations, integral transformations, and mathematical reasoning related to variable substitution in integrals.

Economist08
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Hi
I just cannot understand the following transformation, where [tex]\phi(t)[/tex] is the displacement of an optimal path using standard calculus of variations. All functions are defined between 0 and T. [tex]\phi[/tex] equals zero at 0 and T. r is some discount rate, e it the Euler number, t is time.

[tex]\int^{T}_{0}\theta(y(t))e^{-rt}\int^{t}_{0}\phi(\tau)d\tau]dt[/tex]

This should be equal to

[tex]\int^{T}_{0}\int^{T}_{t}\theta(y(\tau))e^{-r\tau}d\tau]\phi(t)]dt[/tex]

If anybody knows the answer, I would be very happy to get some help here.
Thanks in advance!
 
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Economist08 said:
Hi
I just cannot understand the following transformation, where [tex]\phi(t)[/tex] is the displacement of an optimal path using standard calculus of variations. All functions are defined between 0 and T. [tex]\phi[/tex] equals zero at 0 and T. r is some discount rate, e it the Euler number, t is time.

[tex]\int^{T}_{0}\theta(y(t))e^{-rt}\int^{t}_{0}\phi(\tau)d\tau]dt[/tex]

This should be equal to

[tex]\int^{T}_{0}\int^{T}_{t}\theta(y(\tau))e^{-r\tau}d\tau]\phi(t)]dt[/tex]

If anybody knows the answer, I would be very happy to get some help here.
Thanks in advance!

I think you're being blinded by the detail.

Look carefully and you'll see that the only difference in the two integrands is that t and [tex]\tau[/tex] have been swapped over.

Now, you can always change the names of the variables. For example, ∫f(x)dx = ∫f(y)dy, or ∫∫f(x,y)dxdy = ∫∫f(z,w)dzdw.

In your example, x = w and y = z: ∫∫f(x,y)dxdy = ∫∫f(y,x)dydx.

The only place you have to be careful is in choosing the range of integration.

It is the triangular area 0 < [tex]\tau[/tex] < t; 0 < t < T.

In other words, all pairs of t and [tex]\tau[/tex] with [tex]\tau[/tex] < t < T.

Swapping t and [tex]\tau[/tex] gives: all pairs of t and [tex]\tau[/tex] with t < [tex]\tau[/tex] < T, or:
:smile: t < [tex]\tau[/tex] < T; 0 < t < T. :smile:
 
Thank you so much! Your answer makes perfect sense!
 
Welcome to PF!

Economist08 said:
Thank you so much! Your answer makes perfect sense!

You're very welcome!
Which reminds me, I forgot to say …

:smile: Welcome to PF! :smile:
 

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