Convergence or divergence of the integral $\int\limits_{10}^\infty \frac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)} \,dx$

Question

I want to calculate $ \displaystyle \int_{10}^\infty \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}dx = \lim_{R \to \infty}\int_{10}^R \cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)} \, dx $. I think that it will diverge by bounding it below with $\frac{1}{x}$. So how do we calculate this integral?

Answer 1

Note that

$$\cfrac{e^x}{x^{1 + \frac{1}{x}} (e^x -1)}\sim\frac 1x$$

then the integral diverges by limit comparison test with $\int \frac1x dx$.

Answer 2

Since $x^{-1/x}$ decreases for $x>e$ and $\frac{e^x}{e^x-1}$ decreases for $x>0$, then we have for $x>10$

$$\frac{e^x}{x^{1+1/x}(e^x-1)}> \left(\frac{e^{10}}{10^{1/10}(e^{10}-1)}\right)\,\frac1x$$

By comparison, the integral diverges.