Why does this not diverge when x = inf?

[denjay]

Here is the equation I came across during a proof in my Statistical Mechanics textbook.

\frac{x^3}{3} ln(1-e^{-βx}) = 0

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

Why does this not diverge? Is it an estimation?

 

[Mute]

Here is the equation I came across during a proof in my Statistical Mechanics textbook.

\frac{x^3}{3} ln(1-e^{-βx}) = 0

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

Why does this not diverge? Is it an estimation?

What do you mean by "x is evaluated from \infty to 0"? 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.

 

[Mark44]

Here is the equation I came across during a proof in my Statistical Mechanics textbook.

\frac{x^3}{3} ln(1-e^{-βx}) = 0

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.

where β = \frac{1}{k_{B}T} and x is evaluated from \infty to 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 → ∞.

Why does this not diverge? Is it an estimation?

 

[denjay]

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.

 

[HallsofIvy]

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