Cts approximation, delta function integration, stat mech

[binbagsss]

Homework Statement

Homework Equations

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##?

Many thanks in advance

 

[stevendaryl]

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

Don't you mean for [itex]dx'[/itex] to be inside the integral? If so, you're basically done. For any [itex]x_0[/itex],

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

So we have the particular case [itex]x_0 = E-\frac{1}{2}\bar{h}\omega[/itex]