In the lectures I quoted, without proof, the results

and

which are good for low enough values of . Here is the derivation (and I only recommend the following for those who are particularly good at maths).

To get the chemical potential we use

with and to get the internal energy we have a similar integral, just an extra factor of in the numerator.

Therefore we have to perform integrals of the type

where for and for .

It must be possible to write

You can see that this must be possible because at the denominator of the integrand is equal to unity for and infinite for Let's try and manipulate our integral into this form.

Put . Then we have

Now use a neat trick and write

This leads to

Now we notice that the middle of these three integrals has an upper limit which we can safely set to infinity. You can see this because as the integrand is approximately equal to . Since is only powerlike in , it follows that the contribution to the integral from the large region is suppressed exponentially, i.e. by a factor . Such a term will be smaller than any power of as . Therefore we can safely write

We have succeeded in manipulating the integral into a useful form. The first term is the term we calculated in the lectures whilst the second term can be expanded in a Taylor series about , i.e.

The integral over gives

We can now go ahead and determine the chemical potential and the internal energy.

For the chemical potential

and for the internal energy