# Why does this not diverge when x = inf?

1. Dec 4, 2012

### 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?

2. Dec 4, 2012

### Mute

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.

3. Dec 4, 2012

### 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 → ∞.

4. Dec 4, 2012

### 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.

5. Dec 5, 2012

### HallsofIvy

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