# Pesky change of variables in integral

1. Jul 19, 2012

### w4k4b4lool4

Hi All,

I've managed to confuse myself with a simple change of variables.

I have an integral of the form:
$$I = \int_f^{\infty} dt \int_0^1 ds\, t\, F(t(1-s),ts),$$
where $F(a,b)$ is some well behaved function and $f$ is a positive number. I want to change variables:
$$x = t(1-s), \qquad y = ts,$$
in terms of which the integral reads:
$$I = \int_{x_i}^{x_f} dx \int_{y_i}^{y_f} dy\, F(x,y).$$
I would naively conclude that:
$$x_i=y_i=0,\qquad x_f=y_f=\infty,$$
but this must be wrong because it is independent of f!

My question is: what are the limits of integration in the new variables and why?

Wakabaloola

2. Jul 19, 2012

### mathman

I suggest you do it one step at a time. First y = ts, so dy = tds, while the y limits become 0 and t. Second step, switch integration order, so t is inner integral. Finally let x = t-y, so dx=dt and the x limits will change accordingly.

3. Jul 19, 2012

### w4k4b4lool4

$$I = \int_f^{\infty}dt \int_0^t dy F(t-y,y).$$
How can I then switch the integration order (given that the y integral has t-dependent limits)?

4. Jul 19, 2012

### w4k4b4lool4

OK, I think I figured it out.

The region of integration in the (t,s) variables is the rectangle specified by the four vertices (f,0), (f,1), (L,1) and (L,0), with L->Infinity.

In the new variables (x,y), the region of of integration is a trapezoidal specified by four vertices (f,0), (0,f), (0,L) and (L,0), which (taking L->Infinity) can equivalently be described as the area of the first quadrant of the (x,y) plane less the area associated to a triangle with vertices (0,0), (0,f) and (f,0). That is,
$$I = \int_0^{\infty}dx \int_0^{\infty}dy \,F(x,y) - \int_0^f dx\int_0^{f-x}dy\, F(x,y)$$

5. Jul 19, 2012

### HallsofIvy

Staff Emeritus
You are treating t as a constant here which is not true.

With x= t(1- s), y= st, t= x+ y and s= y/(x+ y).

$$dsdt= \left|\begin{array}{cc}\frac{\partial s}{\partial x} & \frac{\partial s}{\partial y} \\ \frac{\partial t}{\partial x} & \frac{\partial t}{\partial y}\end{array}\right|dxdy= \left|\begin{array}{cc} y & x+ 2y \\ 1 & 1 \end{array}\right|dxdy= -(x+y)dxdy$$

6. Jul 19, 2012

### w4k4b4lool4

$$dx\wedge dy = (dt(1-s)-tds)\wedge (tds+sdt) = (1-s)t dt\wedge ds-st ds\wedge dt = tds\wedge dt$$

7. Jul 19, 2012

### Muphrid

You just made a small mistake with the combining of terms here.

$$(1-s)t \; dt \wedge ds - st \; ds \wedge dt = (t - st + st) \; dt \wedge ds = t \; dt \wedge ds = - t \; ds \wedge dt$$

Nevertheless, you already have the integral in terms of $ds \wedge dt$. You'll have to go backwards to find it in terms of $dx \wedge dy$.

8. Jul 19, 2012

### w4k4b4lool4

what mistake did I make, sorry I don't see it!

9. Jul 19, 2012

### Muphrid

When you multiplied out $(1-s)t \; dt \wedge ds$ you somehow got a term that was $t \; ds \wedge dt$ instead of $t \; dt \wedge ds$. This caused your result to be off by a factor of -1.

10. Jul 19, 2012

### w4k4b4lool4

Oh yes, I see, thanks! (I may be dyslexic)

11. Jul 19, 2012