# Cts approximation, delta function integration, stat mech

1. Mar 25, 2017

### binbagsss

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

So cts approx holds because $\frac{E}{\bar{h}\omega}>>1$

So
$\sum\limits^{\infty}_{n=0}\delta(E-(n+1/2)\bar{h} \omega) \approx \int\limits^{\infty}_{0} dx \delta(E-(x+1/2)\bar{h}\omega)$

Now if I do a substitution $x'=x\bar{h}\omega$ to loose the $\bar{h}\omega$ multiplying the $x$ , $dx'=\bar{h}\omega dx$

I get
$dx' \frac{1}{\bar{h}\omega}\int\limits^{\infty}_{0}\delta(x'-(E-\frac{1}{2}\bar{h}\omega))$

Now, if I denote the region that $x'$ is integrated over by $D$ I get that this is:
$= \frac{1}{\bar{h}\omega}$ if $x'=E-1/2\bar{h}\omega \in D$
$= 0$ if $x'=E-1/2\bar{h}\omega \notin D$

The solution however has:

$= \frac{1}{\bar{h}\omega}$ if $E>1/2\bar{h}\omega$
$= 0$ if $E<1/2\bar{h}\omega$

Excuse me if I'm being stupid but I have no idea how we have converted the requirements of a certain value of $x'$ to lie inside the region of integration or not, which I believe is the definition of the delta function, to inequalities imposed on $E$?

2. Mar 27, 2017

### stevendaryl

Staff Emeritus
Don't you mean for $dx'$ to be inside the integral? If so, you're basically done. For any $x_0$,

$\int_0^{\infty} \delta(x' - x_0) dx' = 0$ if $x_0 < 0$
$\int_0^{\infty} \delta(x' - x_0) dx' = 1$ if $x_0 >0$

So we have the particular case $x_0 = E-\frac{1}{2}\bar{h}\omega$