I'm trying to prove the following integral converge: $$ \int_{0}^{\infty}\frac{e^{-\frac{1}{x}}-1}{x^\frac{2}{3}} $$ since 0 and $\infty$ are the problematic points I've done this: $$ \int_{0}^{\infty}\frac{e^{-\frac{1}{x}}-1}{x^\frac{2}{3}} = \int_{0}^{1}\frac{e^{-\frac{1}{x}}-1}{x^\frac{2}{3}} + \int_{1}^{\infty}\frac{e^{-\frac{1}{x}}-1}{x^\frac{2}{3}} $$
now I know: $$\int_{0}^{1}\frac{e^{-\frac{1}{x}}-1}{x^\frac{2}{3}} \leq \int_{0}^{1}\frac{1}{x^\frac{2}{3}} < \infty$$ because: $$\int_{0}^{1}\frac{1}{x^\alpha } < \infty \iff \alpha <1$$
Now, I'm just having difficulties proving the other part converges, any help is appreciated!
Change variables: $y=1/x$, $dx/x=-dy/y$. Then the integral becomes $$ \int_0^{\infty} \frac{e^{-y}-1}{y^{4/3}} \, dy. $$ Split at $y=1$, and then $$ \int_1^{\infty} \frac{1-e^{-y}}{y^{4/3}} \, dy $$ is bounded by $\int_1^{\infty} \frac{dy}{y^{4/3}}$, which converges. Meanwhile, on $[0,1]$ we have $$ \frac{e-1}{e}y \leqslant 1-e^{-y} \leqslant y $$ (draw a picture, look at the tangent at $0$ and the secant through $0$ and $1$), so the integral over $[0,1]$ is bounded by a multiple of $$ \int_0^1 \frac{y}{y^{4/3}} \, dy = \int_0^1 \frac{dy}{y^{1/3}}, $$ which converges.