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Why does this not diverge when x = inf?

  1. Dec 4, 2012 #1
    Here is the equation I came across during a proof in my Statistical Mechanics textbook.

    [itex]\frac{x^3}{3} ln(1-e^{-βx}) = 0[/itex]

    where [itex]β = \frac{1}{k_{B}T}[/itex] and [itex]x[/itex] is evaluated from [itex]\infty[/itex] to [itex]0[/itex].

    Why does this not diverge? Is it an estimation?
  2. jcsd
  3. Dec 4, 2012 #2


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    Homework Helper

    What do you mean by "[itex]x[/itex] is evaluated from [itex]\infty[/itex] to [itex]0[/itex]"? Just that that is the range of x, i.e., ##x \in [0,\infty)##?

    At any rate, as you take ##x\rightarrow \infty##, ##e^{-\beta x}## becomes small. You can then use the taylor expansion of the logarithm, ##\ln(1+t) \approx t## as ##t\rightarrow 0## to find the leading order behavior as x gets large, and see that the expression will go to zero.

    You could also use L'Hopital's rule to demonstrate this, as the limit is of the indeterminate form ##\infty \cdot 0##. Both approaches will give you the limit, but the first approach also tells you how the function behaves asymptotically as x gets large.
  4. Dec 4, 2012 #3


    Staff: Mentor

    Was this in the form of a limit? That context would make sense with what you said below.

    $$ \lim_{x \to \infty} \frac{x^3}{3} ln(1-e^{-βx}) $$

    The above is the indeterminate form [∞ * 0]. Being indeterminate, you can't tell how it will turn out.

    The thing to do is to write it in a form so that L'Hopital's Rule can be used; namely,
    $$ \frac{1}{3}\lim_{x \to \infty} \frac{ln(1-e^{-βx}) }{x^{-3}} $$

    Now, it's the indeterminate form [0/0], so L'Hopital's can be used. The numerator is approaching 0, as is the denominator.

    Applying L'H, you arrive pretty quickly at a limit value of 0.
    I'm assuming that β > 0. Also, we don't normally evaluate something from ∞
    to 0. We can take the limit, though, as x → ∞.
  5. Dec 4, 2012 #4
    Sorry I probably should have provided more information. That equation is the result of an integral being from 0 to ∞. So, with that equation, setting x = 0 makes the equation zero along with setting x = ∞.

    And the only restriction on β is that β > 0.
  6. Dec 5, 2012 #5


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    Science Advisor

    If you are working with integrals, you should know by now that you can not "set [itex]x= \infty[/itex]". You have to use limits.
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