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Pesky change of variables in integral

  1. Jul 19, 2012 #1
    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?

    Thanks in advance!
  2. jcsd
  3. Jul 19, 2012 #2


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    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.
  4. Jul 19, 2012 #3
    So, first step leads to:
    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)?

    Thanks for your response!
  5. Jul 19, 2012 #4
    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)
  6. Jul 19, 2012 #5


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

    You need, instead, the "Jacobian".
    With x= t(1- s), y= st, t= x+ y and s= y/(x+ y).

    [tex]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[/tex]
  7. Jul 19, 2012 #6
    I find instead:
    dx\wedge dy = (dt(1-s)-tds)\wedge (tds+sdt) = (1-s)t dt\wedge ds-st ds\wedge dt = tds\wedge dt
  8. Jul 19, 2012 #7
    You just made a small mistake with the combining of terms here.

    [tex](1-s)t \; dt \wedge ds - st \; ds \wedge dt = (t - st + st) \; dt \wedge ds = t \; dt \wedge ds = - t \; ds \wedge dt[/tex]

    Nevertheless, you already have the integral in terms of [itex]ds \wedge dt[/itex]. You'll have to go backwards to find it in terms of [itex]dx \wedge dy[/itex].
  9. Jul 19, 2012 #8
    what mistake did I make, sorry I don't see it!
  10. Jul 19, 2012 #9
    When you multiplied out [itex](1-s)t \; dt \wedge ds[/itex] you somehow got a term that was [itex]t \; ds \wedge dt[/itex] instead of [itex]t \; dt \wedge ds[/itex]. This caused your result to be off by a factor of -1.
  11. Jul 19, 2012 #10
    Oh yes, I see, thanks! (I may be dyslexic)
  12. Jul 19, 2012 #11
    Thanks to all who thought about this question!
    I am satisfied with the answer in post n.4.
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