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Why is this integral zero

  1. Aug 18, 2010 #1


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    I've just found that, for all a>0:

    [tex] \int_0^\infty \frac{a (x^2 - 1)^2 - 2 x (x + a)^2}{(x + a)^3 (a x + 1)^3} dx = 0[/tex]

    This can be found by brute force, but there must be a simpler way to show it.
  2. jcsd
  3. Aug 18, 2010 #2


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    How formal does the solution need to be? I have an idea that might work out, but I don't know if it would be exactly what you're looking for.
  4. Aug 18, 2010 #3


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    Id be interested to hear any thoughts you have on it.
  5. Aug 18, 2010 #4
    The integrand happens to be an exact derivative of x*P(x)/Q(x) for some P and Q.
  6. Aug 18, 2010 #5


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    Read This First: Here are my first thoughts. The biggest problem with this is that it's not rigorous and utilizes a lot of 'hand-waving arguments' (and I'm not positive that this hand-waving is even justified here!). Another problem with it is that it's not really all that simple, so even if it can be justified, I don't think that it's much good as a solution. I think that hamster143 outlined a much better solution too. But with that qualified, here was my first thought about the problem.

    Let the function [itex]f[/itex] be defined such that

    [tex]f(x) = \frac{a(x^2-1)^2-2x(x+a)^2}{(x+a)^3(ax+1)^3}[/tex]​

    Since [itex]f[/itex] is continuous on [itex][0,\infty)[/itex], we can apply the second fundamental theorem of calculus to find that ...

    [tex]\int_0^{\infty}f(x)\mathrm{d}x = \lim_{x \to \infty}F(t) - F(0)[/tex]​

    where [itex]F[/itex] is an anti-derivative of [itex]f[/itex]. Therefore, we need only find the values of these anti-derivatives. We'll go about this in an indirect way.

    To find [itex]F(0)[/itex], first define the function [itex]g[/itex] such that [itex]g(x) = a(x^2-1)^2-2x(x+a)^2[/itex]. Next, note that by choosing [itex]x[/itex] small enough, we can find numbers [itex]h,k > 0[/itex] such that the following inequality holds:

    [tex]h[g(x)] \leq f(x) \leq k[g(x)][/tex]​

    Since it's easy to verify that [itex]G(0) = 0[/itex] (neglecting the constant) where [itex]G[/itex] is an anti-derivative of [itex]g[/itex], this suggests that the anti-derivative of [itex]f[/itex] at zero is equal to zero. Therefore, [itex]F(0) = 0[/itex].

    To evaluate the term [itex]\lim_{x \to \infty}F(t)[/itex], we can note that as [itex]x[/itex] becomes arbitrarily large, [itex]f(x)[/itex] tends to something like [itex]h(x) = x^{-2}[/itex] (because the numerator is a polynomial of degree 4 and the denominator is a polynomial of degree 6). Since [itex]lim_{x \to \infty}H(x) = 0[/itex] (once again, neglecting the constant) where [itex]H[/itex] is an anti-derivative of [itex]h[/itex], this suggests that [itex]\lim_{x \to \infty}F(t) = 0[/itex].

    Combining these two results, we find that ...

    [tex]\int_0^{\infty}f(x)\mathrm{d}x = 0[/tex]​
  7. Aug 19, 2010 #6


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    I'm not sure what you mean. Are P(x) and Q(x) supposed to be polynomials? According to mathematica, the indefininte integral contains logs, and is very complicated.


    I'm not sure I understand exactly what you're saying, but I'm skeptical of your method since you don't seem to be using the specific form of the function I gave. For example, if I changed the (ax+1) in the denominator to (ax+2), the integral would no longer be zero, but I'm not sure your argument would be any different. I think you're neglecting constants when you shouldn't be.
  8. Aug 19, 2010 #7


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    That's very likely. If you read the "Read This First" segment of my post, I acknowledged that.
  9. Aug 19, 2010 #8
    Not according to wolfram alpha ...

  10. Aug 19, 2010 #9


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    "Neglecting" the constant? :confused: You make it sound like there's One True Antiderivative which all others are just modifications of!

    Of course, you can choose the constant so that G(0) = 0. And you can choose the corresponding constant so that F(0) = 0.
    (it would be easier to choose F(0) directly)

    Again with the One True Antiderivative thing. You can certainly choose the constant so that [itex]\lim_{x \to \infty}F(t) = 0[/itex].

    Of course, there's no reason those two choices have to be consistent. :frown:
  11. Aug 19, 2010 #10
    A method which doesn't require heavy developments is given in attachment.

    ( Typo in the attachment : remplace equation -Aa=-a by -Aa=a )

    Attached Files:

    Last edited: Aug 19, 2010
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