Partition function from the density of states

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snatchingthepi
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Homework Statement
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Relevant Equations
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I'm given the following density of states

$$ \Omega(E) = \delta(E) + N\delta(E-\Delta) + \theta(E-\Delta)\left(\frac{1}{\Delta}\right)\left(\frac{E}{N\Delta}\right)^N $$

where $ \Delta $ is a positive constant. From here I have to "calculate the canonical partition function as a function of $$ x=\beta\Delta $$ using the incomplex gamma function

$$ \Gamma(n,x) = \int_x^\infty dt e^{-t} t^{n-1} $$

I know this can be solved for a partition function by taking a Laplace transform of the density of states. I can do the first two term very easily, but for the third term

$$ z_{can} = \int_0^\infty \theta(E-\Delta)\left(\frac{1}{\Delta}\right)\left(\frac{E}{N\Delta}\right)^N exp[-\beta E] dE $$

I'm not sure how to go forward from here. I've never seen an integral like this. I am thinking the step function changes the integral lower bound, but I'm kinda strung out so near the end of term, and am not seeing where to go now. Can anyone please help out?
 
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Please tell us exact step where you are stuck. The integral is doable using incomplete gamma functions
 
I am unsure *how* to do this integrla with the incomplete gamma function. My thought hit a dead-end at

$$ z_{can} = \int_0^\infty \theta(E-\Delta)\left(\frac{1}{\Delta}\right)\left(\frac{E}{N\Delta}\right)^N exp[-\beta E] dE $$

$$ z_{can} = \int_\Delta^\infty \left(\frac{1}{\Delta}\right)\left(\frac{E}{N\Delta}\right)^N exp[-\beta E] dE $$

let $$ x=\beta \Delta $$

and for the incomplete gamma function let

$$ t = \frac{-xE}{\Delta}, dt = \frac{-x}{\Delta} dE $$

so

$$ z_{can} = \left(\frac{1}{N \Delta}\right)^N \int_x^\infty dt t^N exp[-t] $$

$$ z_{can} = \left(\frac{1}{N \Delta}\right)^N \Gamma(N+1, x) $$

I'm not convinced of my math in these last few steps.
 
Last edited:
snatchingthepi said:
Homework Statement:: See post
Relevant Equations:: See post

I'm given the following density of states

$$ \Omega(E) = \delta(E) + N\delta(E-\Delta) + \theta(E-\Delta)\left(\frac{1}{\Delta}\right)\left(\frac{E}{N\Delta}\right)^N $$

where $ \Delta $ is a positive constant. From here I have to "calculate the canonical partition function as a function of $$ x=\beta\Delta $$ using the incomplex gamma function

$$ \Gamma(n,x) = \int_x^\infty dt e^{-t} t^{n-1} $$

I know this can be solved for a partition function by taking a Laplace transform of the density of states. I can do the first two term very easily, but for the third term

$$ z_{can} = \int_0^\infty \theta(E-\Delta)\left(\frac{1}{\Delta}\right)\left(\frac{E}{N\Delta}\right)^N exp[-\beta E] dE $$

I'm not sure how to go forward from here. I've never seen an integral like this. I am thinking the step function changes the integral lower bound, but I'm kinda strung out so near the end of term, and am not seeing where to go now. Can anyone please help out?
You could also replace the factor ##E^N## by ##(-1)^N## times the N-th derivative of the exponential with respect to ##\beta##. The integral will be trivial and then you can apply the N-th derivative wrt ##\beta## on the result to get the final answer.
 
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