# Double integral/change of variable

1. Mar 11, 2009

### hnh

Hello, it may look long but if written down-
is really quite short, I have only included a detailed explanation to try and be more clear.
My question is; I am doing a change of variable which doesn't seem to 'fit' the theory of finding the jacobian to make
the transformation. It seems to me that I make a straightforward substitution:

The integral is \int\int_A f'(t) ln( f(x) - f(t) + t - x)dxdt In my area, x <= t so the order cannot be switched.

I make the following change of variable:

Let r=x-f(x) and h(r) = x and let s=t-f(t) and h(s)=t
notice that this gives f(x)=h(r)-r and f(t)=h(s)-s, hence f'(t)dxdt = dxf'(t)dt = h'(r)dr(h'(s) - 1)ds=h'(r)(h'(s)-1)drds.

Now the integral appears to be

\int\int_A' ln(r-s)h'(r)(h'(s)-1)drds

Is this correct-can I do this change of variable without actually doing the jacobian. It seems that I do not actually
have the integral \int\int f(x,y) dxdt in this case to make the transformation with the jacobian so I am wondering
if there is anything wrong with the way I have done this?

2. Mar 12, 2009

### BobbyBear

I think it would work because your r and s variables are not each one dependent on x and t at the same time, so it's like you're doing two one-variable changes, so you really are doing the Jacobian thingy even if you don't realise that you are, if you know what I mean. You have:

$$r(x) \equiv x -f(x) \ \ \rightarrow \ \ x=h(r);$$
$$s(t) \equiv t-f(t) \ \ \rightarrow \ \ t=h(s);$$

$$f(x)=x-r(x) \ \ \rightarrow \ \ f(x(r)) \equiv F(r) = h(r)-r;$$
$$f(t)=t-s(t) \ \ \rightarrow \ \ f(t(s)) \equiv F(s) = h(s)-s;$$

The Jacobian would be:

$$J = det \left( \begin{array}{cc} \frac {\partial x} {\partial r} & \frac {\partial t} {\partial r}\\ \frac {\partial x} {\partial s} & \frac {\partial t} {\partial s} \end{array} \right) = det \left( \begin{array}{cc} h'(r) & 0\\ 0 & h'(s) \end{array} \right) = h'(r) h'(s)$$

so you have:

$$dxdt = J \ dr ds = h'(r) h'(s) dr ds$$

and

$$ln|f(x(r))-f(t(s))+t(s)-x(r)| = ln|s-r|$$

and

$$f'(t) \ = \ \frac {df} {dt} = \left( 1 - \frac {ds} {dt} \right)$$

And so:

$$\iint_\textrm{A} f'(t) \ ln|f(x)-f(t)+t-x|dxdt = \iint_\textrm{A'} \left( 1 - \frac {ds} {dt} \right) ln|s-r|h'(r) h'(s) dr ds= \iint_\textrm{A'} \left( h'(s)- \frac {ds} {dt} \frac {dh} {ds} \right) ln|s-r|h'(r) dr ds$$

$$= \iint_\textrm{A'} \left( h'(s)- \frac {ds} {dt} \frac {dt} {ds} \right) ln|s-r|h'(r) dr ds= \iint_\textrm{A'} \left( h'(s)- 1 \right) ln|s-r|h'(r) dr ds$$

which is what you got so I think it's correct :)

3. Mar 13, 2009

### hnh

Thank you very much-that is a huge help!

4. Mar 14, 2009

### BobbyBear

You're welcome