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Integral from Unsolvable Equation

  1. Jul 22, 2012 #1
    I'm a bit stumped with a problem I have recently seen. Here it is:

    There is a continuous and decreasing function [tex]y(x):[0,1] \to [0,1],\mbox{ }0<a<b[/tex] and [tex]x^a-x^b=y^a-y^b[/tex]
    Prove that [tex]\int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{3ab}[/tex]

    The trivial solution of y=x causes the integral to diverge. Frankly, I'm at a loss on how to approach this problem. Clearly you cannot solve for y in general as a and b can also take on non-integer values.
     
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  3. Jul 22, 2012 #2

    Besides not being very-very clear, you seem to have given an incorrect integral:
    [tex]\int_0^1\frac{\log y}{x}\,dx=\log y\int_0^1\frac{1}{x}\,dx[/tex]

    and that improper integral diverges...perhaps you meant:

    [tex]\,\,1)\,\,\,\forall\,\,0<a<b<1\,\,\,,\,\,x^a-x^b=y^a-y^b[/tex]

    [tex]\,\,2)\,\,\,\int_0^1 \log\frac{y}{x}\,dx\,\,?\,[/tex]

    Anyway, the above integral is an ugly beast...please clarify this.

    DonAntonio
     
  4. Jul 22, 2012 #3
    That's how the problem was presented. Also, y is a function of x, not a constant. As such, it cannot be removed from the integral.

    I have implicitly plotted [tex]x^a-x^b=y^a-y^b[/tex] for various values of a and b and I can see the two solutions for y. One is x, and the other fits the definition of y(x) given in my original post.
     
    Last edited: Jul 22, 2012
  5. Jul 22, 2012 #4
    In fact, it almost looks like a circle. Putting in [tex]y=\sqrt{1-x^{2}}[/tex] into the integral produces [tex]\int_{0}^{1} \frac{\ln y}{x} dx=-\frac{\pi ^2}{24}[/tex]
     
  6. Jul 22, 2012 #5


    Oh, I see now...and those values of [itex]\,0<a<b\,[/itex] are constant and fixed, uh? Sorry, I don't have the faintest idea, though

    I tried to make a substitution [tex]bx+(1-x)a=u\Longrightarrow dx=\frac{du}{b-a}[/tex] to change the integral's limits so that [itex]\,a,b\,[/itex] get into play, but I can't see how to get them as powers of neither [itex]\,x,y\,[/itex]...good luck!

    DonAntonio
     
  7. Jul 22, 2012 #6

    What is what looks "almost" as a circle??

    DonAntonio
     
  8. Jul 22, 2012 #7
    I said in a previous post:
    I'm talking about the non-trivial solution for y. It's almost a circle.
     
  9. Jul 22, 2012 #8

    PAllen

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    Just to clarify the problem, I worked the trivial case of a=1, b=2. Then you get either y=x, for which the improper integral diverges, or y=(1-x) for which the improper integral is π^2/6, consistent with the stated general formula. I don't have any suggestion for where to go from there ... just that the problem as stated seems plausible.
     
  10. Jul 22, 2012 #9
    It seems unlikely that there will be a closed form solution for y since a and b can take on non-integer values as well. However, there may be a series representation of the logarithm of y. That can be integrated.
     
  11. Jul 22, 2012 #10

    PAllen

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    Where's Ramanujan when you need him ...
     
  12. Jul 23, 2012 #11

    haruspex

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    Change of variable, x = eu+v, y = eu-v. Equation becomes
    e(b-a)u = sinh(av)/sinh(bv)
    Meanwhile, by symmetry about the u-axis, the integral becomes
    I = ∫0(v-u).dv = ∫0((av - ln(sinh(av)) + bv - ln(sinh(bv)))/(b-a))dv
    0(av - ln(sinh(av)))dv = ∫0(v - ln(sinh(v)))dv/a
    I = ∫0(v - ln(sinh(v)))dv/ab
    v - ln(sinh(v)) = -ln(1-e-2v) = Ʃe-2vn/n
    I = Ʃ[-e-2vn/2n2]v=0/ab
    = π2/12ab
    Can anyone see where I lost the factor of 4?
     
  13. Jul 23, 2012 #12

    Mute

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    How did you get to the bolded line? It looks to me like you're missing a factor from the change of variables, [itex]dx/x = dv\left[du(v)/dv + 1\right][/itex]. Where did that go?
     
  14. Jul 23, 2012 #13

    haruspex

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    It's easiest in two stages. First x = er, y = es. The integral becomes ∫-∞0s.dr.
    Next, r = u+v, s = u-v. The tricky part here is the integral. It's best to look at it graphically. In the r-s plane, the curve is in the -/- quadrant, asymptotic to both axes. It's symmetrical about the line r=s (the u-axis). We want the area between the curve and the r-s axes. Cut the area with the line r=s, and fix on the half of the area touching the r axis.
    At a given v, the length of the area in the u direction is v-u. (I think I might have reversed the sign of u somewhere; it looks like this should be u+v, but v-u works.) So the (half) integral is ∫(v-u)dv. OTOH, the switch to u, v has changed the scale: du.dv = 2dr.ds. So I think ∫(v-u)dv is actually the whole integral. But if I have that wrong it might explain one or two factors of 2.
     
  15. Jul 23, 2012 #14
    Is that substitution of y valid as it's an unknown function of x? What is u and v?
     
  16. Jul 23, 2012 #15

    haruspex

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    It's just a change of variables. In graphical terms, a rotation through 45 degrees, but with a bit of scaling too. (To avoid the scaling it would be (u+v)/√2, etc.)
     
  17. Jul 24, 2012 #16

    PAllen

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    I think there are several things wrong with this, as suggested by Mute. However, it looks promising. I suggest working it out with the trivial case I gave (a=1,b=2), to clarify the issues. If I apply my understanding of what you suggest above to this case, it leads to nonsense. Meanwhile, directly dealing with:

    y=(1-x), you get polylog(2,x) for the antiderivative, and this is zero for zero and π^2/6 for 1, giving the right answer. Working out your suggestion for this simple case should track down issues you (I think) have with limits of integration, dropped terms, etc.
     
    Last edited: Jul 24, 2012
  18. Jul 24, 2012 #17

    haruspex

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    Then I haven't explained it well enough. How does it lead to nonsense?
     
  19. Jul 24, 2012 #18

    PAllen

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    For starters, the limits of integration you propose are wrong for this case after your substitution. I get that they should be -∞ to +∞, which could explain one factor of 2. But then, I may be misunderstanding your method.

    What would be helpful (to make sure we are on the same page) is to outline your approach in detail using the simple case. We know that direct computation of this simple case produces the right answer. Applying your method to it should help track down the issues.
     
  20. Jul 24, 2012 #19
    Haruspex, could you use [tex] so we can better see what you're doing? It'll improve readability.
     
  21. Jul 24, 2012 #20

    haruspex

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    The first substitution (x = exp(r)) makes the range r = -∞ to 0. The area now sits in the third quadrant and the curve resembles 1/r. I cut that in half at its axis of symmetry, r = s, and consider only the half adjacent to the negative r-axis. With the second substitution, v = r - s, the range becomes v = -∞ to 0 (I think I had v reversed in my head, so probably had a wrong sign there). The area of interest is now bounded by the curve, the u-axis, and by the line u = v (the r-axis). Hence the height, for purposes of integration, is v-u (both being negative).
     
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