The gamma function is defined by $$\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} dt $$ where $x > 0$. Show that $\Gamma(x) \sim \sqrt{2 \pi} e^{-x}x^{x-\frac12}$.
$\sim$ denotes that the ratio between the left and the right side tends to $1$.
I think that it is equivalent to showing
$$ \lim_{x \to \infty} \frac1{\sqrt{2\pi}} \int_0^\infty t^{x-1}e^{-t+x} x^{\frac12 -x} dt =1.$$
This means that the limit of this integral tends to $\sqrt{2\pi}$, but I don't know how to show this.
I would appreciate if you give some help.
It suffices to show that
$$ \Gamma(x+1) = x\Gamma(x) \sim \sqrt{2\pi} \, x^{x+\frac{1}{2}}e^{-x}. $$
Substitute $t = x + \sqrt{x}s$ and define
$$ f_x(s) := \left(1 + \frac{s}{\sqrt{x}}\right)_+^{x} e^{-\sqrt{x}s}, $$
where $x_+ := \max\{0, x\}$ is the positive part of $x$. Then the integral defining $\Gamma(x+1)$ boils down to:
\begin{align*} \Gamma(x+1) &= \int_{-\sqrt{x}}^{\infty} \left(x + \sqrt{x}s\right)^{x} e^{-(x + \sqrt{x}s)} \sqrt{x} \, \mathrm{d}s \\ &= x^{x+\frac{1}{2}}e^{-x} \int_{-\sqrt{x}}^{\infty} \left(1 + \frac{s}{\sqrt{x}}\right)^{x} e^{-\sqrt{x}s} \, \mathrm{d}s \\ &= x^{x+\frac{1}{2}}e^{-x} \int_{-\infty}^{\infty} f_x(s) \, \mathrm{d}s, \end{align*}
Now we note the following observations:
$\displaystyle \lim_{x\to\infty} f_x(s) = e^{-\frac{s^2}{2}} $
It is easy to prove that $(1 + x)_+ \leq e^{x - \frac{x^2}{2(1+x_+)}}$ for all $x \in \mathbb{R}$. Using this, we get
$$ 0 \leq f_x(s) \leq e^{-\frac{s^2}{2(1+s_+)}} \qquad \text{for all} \quad x \geq 1, \ s \in \mathbb{R}.$$
So, by the dominated convergence theorem, we can interchange the order of limit and integration to get:
$$ \lim_{x\to\infty}\int_{-\infty}^{\infty} f_x(s) \, \mathrm{d}s = \int_{-\infty}^{\infty} \lim_{x\to\infty}f_x(s) \, \mathrm{d}s = \int_{-\infty}^{\infty} e^{-\frac{s^2}{2}} \, \mathrm{d}s = \sqrt{2\pi} $$
Therefore the desired conclusion follows.