I need to find if $$\int_0^\infty \frac {e^{-x}-1} {\sqrt x^3}dx$$ converges.
Let $$f(x) = \int_0^\infty \frac {1- e^{-x}} {\sqrt x^3}dx \ge 0$$
$$\int_0^\infty \frac {1-e^{-x}} {\sqrt x^3}dx = \int_0^1 \frac {1- e^{-x}} {\sqrt x^3}dx + \int_1^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$$
The integral $$\int_1^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$$ converges by applying the comparison test: $$\lim_{x \to \infty}\frac {\frac {1-e^{-x}} {\sqrt {x^3}}} {\frac 1 {\sqrt {x^3}}} = \lim_{x \to \infty}1- e^{-x}=1$$
so $f(x)$ converges iff $$\int_1^\infty \frac 1 {x^{\frac 3 2}}dx$$
converges.
the problem i have is with $\int_0^1 f(x)dx$
i tried to use the same comparison test but got limit 0, so divergence of $$\int_0^1 \frac 1 {x^{\frac 3 2}}dx$$ doesn't indicate that $\int_0^1 f(x)dx$ diverges.
how do i handle $\int_0^1 f(x)dx$ ?