Double Integration change of variables

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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!