Helmholtz free energy from internal energy formula I have just encountered this formula for obtaining the Helmholtz free energy $F(T,V)$ of a system from the internal energy $U(T,V)$:
$$F(T,V)=U(0,V)-T\int_0^TdT'\dfrac{U(T',V)-U(0,V)}{T'^2}.$$
The book I got it from says you can prove it knowing that
$$U(T,V)=-T^2\left[ \dfrac{\partial}{\partial T} \left( \dfrac{F(T,V)}{T} \right) \right]_V$$
and
$$F(T,V)=U(T,V)+T\left(\dfrac{\partial F}{\partial T}\right)_V$$
by employing the third law of thermodynamics.
I already knew the formulas above, but I could not prove the formula. If I try to integrate the second formula I don't know how to deal with having to integrate from zero to $T$ and the temperature being at the denominator. Can someone help me?
 A: 
I have just encountered this formula...$$F(T,V)=U(0,V)-T\int_0^TdT'\dfrac{U(T',V)-U(0,V)}{T'^2}.\tag{A}$$


The book I got it from says you can prove it knowing that
$$U(T,V)=-T^2\left[ \dfrac{\partial}{\partial T} \left( \dfrac{F(T,V)}{T} \right) \right]_V\tag{B}$$
and
$$F(T,V)=U(T,V)+T\left(\dfrac{\partial F}{\partial T}\right)_V\tag{C}$$


Can someone help me?

Instead of considering $F(T, V)$, you can consider:
$$
\Delta F(T, V) \equiv F(T, V) - F(0, V) = F(T,V) - U(0,V)\;.
$$
The utility of considering $\Delta F$ instead of $F$ is that it is easy to show that
$$
\lim_{T\to 0}\frac{\Delta F}{T} \to -S(T=0)\;,
$$
which is finite by the third law of thermodynamics. The above limit is calculated using the usual rule for an indefinite ratio (where both the numerator and denominator go to zero):
$$
\lim_{T\to 0}\frac{\Delta F}{T} = \lim_{T\to 0}\frac{\frac{\partial \Delta F}{\partial T}}{\frac{\partial T}{\partial T}}
$$
$$
=\lim_{T\to 0}\frac{\frac{\partial \Delta F}{\partial T}}{1}
=\lim_{T\to 0}\frac{\frac{\partial F}{\partial T}}{1}
=\lim_{T\to 0}\frac{\partial F}{\partial T} = -S(0)\;.
$$
The above limit can also be evaluated by expanding in a power series like:
$$
\Delta F \equiv F(T, V) - F(0, V) \approx T\frac{\partial F}{\partial T} + O(T^2)
$$
The above limit can also be evaluated by recognizing that it is just the definition of the derivative of $F$ at $T=0$:
$$
\lim_{T\to 0}\frac{\left(F(T,V) - F(0, V)\right)}{\left(T - 0\right)}\equiv \left.\frac{\partial F}{\partial T}\right|_0
$$
So, anyways, using $\lim_{T\to 0}\frac{\Delta F}{T} \to -S(T=0)$, the lower limit at $T=0$ is not a problem.
You can further show, using Eq (B) above, that:
$$
U(T, V) - U(0, V) = -T^2\frac{\partial}{\partial T} \left(\frac{\Delta F}{T}\right)\;. \tag{1}
$$
Rearranging my Eq (1) above gives:
$$
\frac{-1}{T^2}\left(U(T, V) - U(0, V)\right) = \frac{\partial}{\partial T} \left(\frac{\Delta F}{T}\right)\;.\tag{2}
$$
Integrating both sides from $0$ to $T$, I find:
$$
F(T,V) = U(0, V) - TS(0,V) - T\int_{0}^T dT'\frac{U(T',V) - U(0,V)}{T'^2}\;,
$$
which is the same expression as OP's Eq. (A) except for the addition of the term $TS(0, V)$.
If there is a unique ground state then $S(0, V) = 0$ and the two expressions agree, otherwise $S(0, V) = \text{log}(g_0)$, where $g_0$ is the degeneracy of the ground state.
Not sure if OP needs to use $S(0, V) = 0$, or if there is an error in my calculation somewhere, or if there is some other issue.
