Consider $$\Gamma(a+1) = \int_{0}^{\infty}e^{-x} x^a dx = a!$$ Let $$h(u) = (u-1) - \ln(u)$$
then we can write $$e^{-x} x^a = e^{-a+a\ln(a)}e^{-a~h(x/a)}$$ Expand $h(u)$ in Taylor series at $u=1$ $$h(u) \approx \frac{(u-1)^2}{2}$$ Then we can calculate the integral $$ \int_{0}^{\infty}e^{-x} x^a dx = \int_{0}^{\infty} e^{-a+a\ln(a)}e^{-\frac{a}{2}(\frac{x}{a}-1)^2} dx\\ = a~e^{-a+a\ln(a)} \int_{0}^{\infty} e^{-\frac{a}{2}(u-1)^2} du\\ \approx a~e^{-a+a\ln(a)} \sqrt{\frac{2 \pi}{a}}=\sqrt{2\pi a} \left( \frac{a}{e} \right)^a $$ Note that the last integral is gausian integral and when $a \gg 1$ all the area under the curve is concentrated at $u=1$.
However, my question is about approximation $h(u)$ only with leading term,(as we are integrating in higher values of $u$ as well). Why is this valid approximation? Could you show that $$ \lim_{a\rightarrow\infty} \frac{I(a) - \tilde{I}(a)}{I(a)}=0$$ where $\tilde{I}(a)$ is the approximated integral, $I(x)$ without approximation. Or give another guarantees for convergence.
And also, is this general method for aproximating integrals? If so, what is the name of this method, and/or where can I read about it?
Thank you.