Double Integration change of variables

[Economist08]

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

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

This should be equal to

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

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

 

[tiny-tim]

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

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

This should be equal to

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

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 \tau 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 < \tau < t; 0 < t < T.

In other words, all pairs of t and \tau with \tau < t < T.

Swapping t and \tau gives: all pairs of t and \tau with t < \tau < T, or:

t < \tau < T; 0 < t < T. ​

 

[Economist08]

Thank you so much! Your answer makes perfect sense!

 

[tiny-tim]

Welcome to PF!

Thank you so much! Your answer makes perfect sense!

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

Welcome to PF! ​