Showing that $ \int_{10}^\infty \frac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx$ diverge

Question

Is it true that $ \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx$ diverge since $ \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx > \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}}e^x }dx=\displaystyle \int_{10}^\infty \cfrac{1}{x^{1 + \frac{1}{x}} }dx$ and $\displaystyle \int_{10}^\infty \cfrac{1}{x^{1 + \frac{1}{x}} }dx$ diverges by $p$-test?

Answer 1

We have

$$\lim_{x\to+\infty}\Bigl((1+\frac{1}{x})\ln (x)-\ln (x)\Bigr)=0$$ then $$ x^{1+\frac{1}{x}} \sim x \; (x\to +\infty) $$

on the other hand $$e^x\sim (e^x-1) \; (x \to +\infty) $$

thus

$$f (x)\sim \frac {1}{x} \;(x\to+\infty) $$

so $$\int^{+\infty}f (x)dx \text { diverges} $$