I'm given this integral $$ \int_{0}^{\infty}\frac{1-e^{-x}}{x\sqrt{x}}dx $$ I need to tell if this integral diverges, converges or absolutely converges
My efforts:
Because $0$ causes problems, split the integral like $$ \int_{0}^{1}\frac{1-e^{-x}}{x\sqrt{x}}dx \int_{1}^{\infty}\frac{1-e^{-x}}{x\sqrt{x}}dx. $$
For the integral $1$ to $\infty$:
$$
f(x)=1-e^{-x},g(x)=\frac{1}{x\sqrt{x}}
$$
and using Dirichlet test, is it possible?
and for the integral $0$ to $1$, use comparison test, and choose $g(x)$ as $$ g(x)=\frac{1}{x^{\frac{3}{2}}} $$
I hope my question is clear.
Notice that the integrand is non-negative so there's no need to investigate absolute convergence separately. We have for $x>0$: $$0\leq \frac{1-e^{-x}}{x\sqrt{x}}\leq\frac{1}{\sqrt{x}}\\ 0\leq \frac{1-e^{-x}}{x\sqrt{x}}\leq\frac{1}{x\sqrt{x}} $$ The second inequality follows from $1-e^{-x}\leq 1$. The first one follows from $e^{t}\geq t+1$ with $t=-x$ which gives us $1-e^{-x}\leq x$. Finally, since the integrals $\int_0^{1}\frac{1}{\sqrt{x}}\,\mathrm{d}x$ and $\int_1^{\infty}\frac{1}{x\sqrt{x}}\,\mathrm{d}x$ converge, we can conclude that the two integrals you mentioned also converge.
Regarding your question about Dirichlet's test, with this choice of functions, we'd need the antiderivative $F(x)=\int_1^x\left(1-e^{-t}\right)\,\mathrm{d}t$ to be uniformly bounded for $x\ge 1$, but $F(x)=x+e^{-x}-(1+e^{-1})$ is clearly unbounded as $x\to\infty$.