We fix $x > 0$ and for all natural integers $n$, we define $F_{n}(x)$ by:
$$F_{n}(x) = \ln\left(\frac{n! \, n^{x+1}}{(x+1)(x+2) \ldots (x+n+1)}\right)$$
and
$$G_{n}(x) = \ln n! + (x+1)\ln n - \left(x+n+\frac{3}{2}\right)\ln(x+n+1) + n+1 + \left(x+\frac{1}{2}\right)\ln x$$
\textbf{V.C.1)} Using Stirling's formula, show that:
$$\lim_{n \rightarrow +\infty} G_{n}(x) = \left(x+\frac{1}{2}\right)\ln x - x + \ln\sqrt{2\pi}$$
\textbf{V.C.2)} Deduce that:
$$\ln \Gamma(x+1) = \left(x+\frac{1}{2}\right)\ln x - x + \ln\sqrt{2\pi} - \int_{0}^{+\infty} \frac{h(u)}{u+x} du$$