I am trying to solve the following problem:
Given the definition of $I_N(u)$ as
\begin{equation}
I_{N}(u) \equiv \frac{1}{2\pi} \int^{\infty}_{0} ds \int^{2\pi}_{0} d\phi\,e^{-s} \left(1 + us + 2\sqrt{us} \cos\phi \right)^{N}
\end{equation}
use integration methods to show that
\begin{equation}
\frac{\partial}{\partial u} I_{N}(u) = N^2 I_{N-1}(u).
\end{equation}
Motivation/Context
The integral identity comes up in a statistical mechanics problem. I am trying to find multiple ways to derive it. A series-based demonstration is presented below.
Mathematical Background
Using the series definition of the Legendre polynomial \begin{equation} P_N(x) = \left(\frac{x-1}{2}\right)^N \sum_{k=0}^{N}\binom{N}{k}^2 \left(\frac{x+1}{x-1}\right)^k, \end{equation} and its integral representation \begin{equation} P_N(x) = \int^{1}_{0} dt \, \left(x+\sqrt{x^2-1}\cos(2 \pi t)\right)^N, \end{equation} it is possible to show that \begin{equation} \sum_{k=0}^N \binom{N}{k}^2 k!\,u^k = \frac{1}{2\pi} \int^{\infty}_{0} ds \int^{2\pi}_{0} d\phi\,e^{-s} \left(1 + us + 2\sqrt{us} \cos\phi \right)^{N}. \end{equation} Differentiating this series we can show \begin{equation} \frac{\partial}{\partial u} \sum_{k=0}^N \binom{N}{k}^2 k! \,u^k = N^2 \sum_{k=0}^{N-1} \binom{N-1}{k}^2 k!\,u^{k}, \end{equation} which provides a series-based proof of the desired identity. I am seeking an integration-based proof.
If we consider $$I_{N}(u) = \frac{1}{2\pi} \int^{\infty}_{0}e^{-s}\, ds \int^{2\pi}_{0} \left(1 + us + 2\sqrt{us} \cos\phi \right)^{N}\, d\phi$$ we have $$\frac 1 \pi \int^{2\pi}_{0} \left(1 + us + 2\sqrt{us} \cos\phi \right)^{N}\, d\phi=$$ $$\left(s u-2 \sqrt{s u}+1\right)^N \, _2F_1\left(\frac{1}{2},-N;1;-\frac{4 \sqrt{s u}}{s u-2 \sqrt{s u}+1}\right)+$$ $$\left(s u+2 \sqrt{s u}+1\right)^N \, _2F_1\left(\frac{1}{2},-N;1;\frac{4 \sqrt{s u}}{s u+2 \sqrt{s u}+1}\right)$$ (provided that $\Re\left(s u-2 \sqrt{s u}\right)>-1 $) which is not the most pleasant thing to integrate again with respect to $s$ for a general $N$.
However, for specific values of $N$, the integration provides quite simple results $$\left( \begin{array}{cc} N & I_{N}(u) \\ 1 & u+1 \\ 2 & 2 u^2+4 u+1 \\ 3 & 6 u^3+18 u^2+9 u+1 \\ 4 & 24 u^4+96 u^3+72 u^2+16 u+1 \\ 5 & 120 u^5+600 u^4+600 u^3+200 u^2+25 u+1 \\ 6 & 720 u^6+4320 u^5+5400 u^4+2400 u^3+450 u^2+36 u+1 \\ 7 & 5040 u^7+35280 u^6+52920 u^5+29400 u^4+7350 u^3+882 u^2+49 u+1 \end{array} \right)$$ where clear patterns seem to appear. In fact, they are $$I_N(u)=\sum_{k=0}^N k! \, \binom{N}{k}^2 u^k=(-u)^N\, U\left(-N,1,-\frac{1}{u}\right)$$