Minimum value of modified Gamma functions

Question

I have a question on the Gamma function. I would like to show that the function $$ G(x) := \frac{\Gamma(x)}{x} = \frac{1}{x} \int_{0}^{\infty} t^{x - 1} e^{- t} dt \quad (x > 0) $$ attains a minimum value on $\mathbb{R}_{+} := \{ x \in \mathbb{R} \mid x > 0\}$. By drawing the graph of $G(x)$, I could confirm that $G(x)$ attains the minimum value at $x_{0}$ such that $1.5 \leq x_{0} \leq 2.5$. However, I cannnot see how we can prove it. I don't need a minimum value, but I would like to prove $G(x)$ attains the minimum on $\mathbb{R}_{+}$. I am happy someone could help me. Thank you.

Answer 1

On $x>0$, $f(x)=\frac{\Gamma(x)}{x}$ is continuous, and $$\lim_{x\to\infty}=\infty,\\\lim_{x\to0+}=\infty$$ so $f$ must attain a minimum on $x>0$.