# I don't understand an approximation in an expression in stat. mech.

1. Oct 5, 2014

### fluidistic

1. The problem statement, all variables and given/known data
Hello guys, I fail to understand a mathematical approximation I see in a solved exercise.
The guy reached a partition function of $Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]$ and he wants to analyze the case $T>> \omicron$.
He states that with the change of variables $x=l(l+1)\frac{\omicron}{T}$, $Z\approx \frac{T}{\omicron} \int _0^\infty e^{-x}dx$.
I really don't understand this last step.
2. Relevant equations
Sum becomes integral. The change of variables.

3. The attempt at a solution
Making the change of variables, I understand that the sum transforms into an integral and I also understand why the limits of the integral are 0 and infinity (because l goes from 0 to infinity and thus x too).
I am unable to perform the change of variables and get rid of l's outside the exponential.
I'd appreciate if someone shed some light. Thanks.

2. Oct 5, 2014

### vela

Staff Emeritus
I think the l's get sucked into what becomes dx since the derivative of l(l+1) is 2l+1.

3. Oct 5, 2014

### RUber

Notice that once you insert the substitution, $Z=\sum_{l=0}^\infty (2l+1) \exp \left [ -l(l+1) \frac{\omicron}{T} \right ]$, $=\sum_{l=0}^\infty (2l+1) \exp [-x]$.
Then see that $dx=(2l+1)\frac{\omicron}{T}$.
So, $2l+1=dx\frac{T}{\omicron}$.
Since $T>>o$ the integral is a good approximation, since it is the limit as $l(l+1)o/T \to 0$.
Putting this all together, $Z = \int_{l=0}^\infty \frac{T}{\omicron} \exp [-x] dx$.

4. Oct 6, 2014

### fluidistic

Thank you guys. I missed the dx part indeed.