Leibniz rule of integration applied to Debye heat capacity The quantity $C_V = \dfrac{\partial U}{\partial T}$ is called the heat capacity in physics, where $U$ is a function of temperature $T$ and other variables. For the case of the Debye's model of a solid, the function $U$ is an energy, and can be shown to be $$U = 9Nk\frac{1}{T_D^3}\int_0^{x_D}\frac{x^3T^4}{e^x-1}dx,$$ where $N, k, T_D$ are constants and $x_D = \hbar\omega_D/(kT),$ where $\hbar,\omega_D$ are also constants. If I apply Leibniz integral rule I get that the derivative of the integral with respect to $T$ is $$ \int_0^{x_D}\frac{4T^3 x^3}{e^x - 1}dx + \frac{T^4x_D^3}{e^{x_D}-1}\dfrac{\partial x_D}{\partial T}.$$ However, the result that is on the textbooks is $$\int_0^{x_D}\frac{T^3x^4e^x}{(e^x-1)^2}dx.$$ I am struggling in getting there, how can I reach the expected result? Thanks for your help.

Both results are indeed consistent, it all comes down to an integration by parts,

Starting from the textbook result, we can integrate by parts,

$$\int_0^{x_D}\frac{T^3x^4e^x}{(e^x-1)^2}dx = \frac{-x^4 T^3}{(e^x-1)}\Big|_0^{x_D}+\int_0^{x_D}\frac{4T^3 x^3}{e^x - 1}dx.$$

Now we rewrite the first term,

$$\frac{-x^4 T^3}{(e^x-1)}\Big|_0^{x_D} = \frac{-x_D^4 T^3}{(e^{x_D}-1)} = \frac{x_D^3 T^4}{(e^{x_D}-1)} \cdot \frac{-x_D}{T} = \frac{x_D^3 T^4}{(e^{x_D}-1)} \dfrac{\partial x_D}{\partial T}. $$

Then the first equation becomes,

$$\int_0^{x_D}\frac{T^3x^4e^x}{(e^x-1)^2}dx = \frac{x_D^3 T^4}{(e^{x_D}-1)} \dfrac{\partial x_D}{\partial T} +\int_0^{x_D}\frac{4T^3 x^3}{e^x - 1}dx,$$

which is the desired result.