# Double Integration change of variables

1. Feb 29, 2008

### 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.

2. Feb 29, 2008

### tiny-tim

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.

3. Feb 29, 2008