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How to prove an inequality for a definite exponential integral

  1. May 31, 2012 #1
    Hello gurus,

    I've been trying to prove the following inequality for several days:

    [itex]\int_1^\infty \frac{\exp\left(-\frac{(x-1)^2}{2a^2}\right)}{x}dx > \ln(1+a)\quad \forall a>0. [/itex]

    I've demonstrated by simulations that this inequality holds. I‘ve also proved that this inequality holds for large [itex]a[/itex]. But, proving [itex]\forall a[/itex] exhausted me...

    Who can help? Many thanks!

    Paola
     
    Last edited: May 31, 2012
  2. jcsd
  3. Jun 2, 2012 #2

    haruspex

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    A possibility is to split the range of integration at 1+a. The 1 to 1+a part then looks to be a bit less than ln(1+a)
    [itex]\int_1^{a+1} \frac{\exp\left(-\frac{(x-1)^2}{2a^2}\right)}{x}dx = \int_0^a \frac{\exp\left(-\frac{x^2}{2a^2}\right)}{1+x}dx [/itex]
    We can then try to top it up to reach ln(1+a) by folding back the rest of the range, e.g. x = (a+1)2/y:
    [itex]\int_{a+1}^\infty \frac{\exp\left(-\frac{(x-1)^2}{2a^2}\right)}{x}dx = [/itex][itex]-\int_{a+1}^0 \frac{\exp\left(-\frac{((a+1)^{2}/y-1)^2}{2a^2}\right)}{y}dy [/itex]
    So if you could show that
    [itex]\exp\left(-\frac{x^2}{2a^2}\right) + \exp\left(-\frac{((a+1)^{2}/(x+1)-1)^2}{2a^2}\right)[/itex] >= 1
    your result would follow.
    But note that in the folding back the integration range became x = -1 to a. So we've wasted some in the -1 to 0 range. That might be critical. If so, a possible way to fix that is to break it further into ranges a to a2, to a3 and so on, folding each past a back onto the target range.
     
  4. Jun 4, 2012 #3

    haruspex

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    .. which seems to be true for a < 2, but it breaks much above that. Asymptotically, looks like it goes as low as 0.95 when x = 0.34*a.
     
  5. Jun 4, 2012 #4
    Let f(a) be the integral on the left. We can write:

    $$ f(a) = \int_0^\infty \frac{1}{x+1} \exp(\frac{-x^2}{2a^2})\, dx.$$

    Let g(a)= ln(1+a). f(0) = g(0) = 0. For f you can sub y = ax and take the limit as a goes to zero.

    [itex] \frac{1}{a\sqrt{\pi/2}}\int_0^\infty h(x) \exp(\frac{-x^2}{2a^2})\,dx [/itex] goes to h(0) as a goes down to zero. This shows that [itex] f'(0) = \sqrt{\pi/2} > g'(0)[/itex].

    This proves the inequality at least in a neighborhood of 0.

    [EDIT] fixed confusing notation that used g(x) in two different senses in the same line.
     
    Last edited: Jun 4, 2012
  6. Jun 4, 2012 #5
    Cool problem. I think I have it now, but I do not guarantee the nonexistence of errors or vastly simpler proofs....

    Notation:

    [itex] g(x) = \frac{1}{1+x} [/itex]

    Notice that [itex] \log(a+1) = \int_0^a g(x)\, dx = \int_0^\infty g(x) \chi_{[0,a]}(x)\, dx [/itex].

    So we want to prove:
    $$ \int_0^\infty g(x) e^{-x^2/2a^2}\, dx > \int_0^\infty g(x) \chi_{[0,a]}(x)\, dx.$$

    Subtract from both sides to reformulate this inequality as [itex] \int_0^\infty g(x)H(x/a)\, dx > 0 [/itex]. By a change of variables, we get
    $$ \int_0^\infty ag(ax) H(x)\, dx >0 ,$$
    where [itex] H(x) = e^{-x^2/2} - \chi_{[0,1]}(x) [/itex]. By setting b= 1/a, we get:
    $$ \int_0^\infty \frac{1}{x+b} H(x)\, dx >0 .$$

    I will just point out that if b=0, this integral is well defined and positive as can be verified through Wolfram alpha or some analytic technique. We break up the integral into two parts and wish to prove that for any positive b,
    $$ \int_0^1 \frac{1}{x+b} (1-e^{-x^2/2})\, dx < \int_1^\infty \frac{1}{x+b}e^{-x^2/2}\, dx.$$

    The left integrand is less than [itex] \frac{1}{x}\frac{1}{1+b} (1-e^{-x^2/2}).[/itex]
    The right integrand is greater than [itex] \frac{1}{x}\frac{1}{1+b} e^{-x^2/2}.[/itex] Therefore the inequality is true if
    $$ \frac{1}{1+b} \int_0^1 \frac{1}{x}(1-e^{-x^2/2})\, dx < \frac{1}{1+b}\int_1^\infty \frac{1}{x}e^{-x^2/2}\, dx.$$

    So it is reduced to a single integral inequality which can be checked explicitly. In fact, this is the same as the inequality above that said was true by Wolfram Alpha.
     
  7. Jun 5, 2012 #6

    haruspex

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    Very neat, Vargo. Looks right to me.
     
  8. Jun 5, 2012 #7
    haruspex and Vargo, thank you so much!

    Vargo, how beautiful your proof is! Thanks a lot!!
     
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