<s>[INST] Convergence of an improper integral $\int_0^\infty \frac {1-e^{-x}} {\sqrt x^3}dx$ 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 $\int_0^1 f(x)dx$ ?
 [/INST] Given
$$
\int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx.
$$
Use substitution $x = y^2$, then we obtain
$$
\begin{eqnarray}
\int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx &=& 2 \int_0^\infty \frac{ \exp(-y^2) - 1 }{ y^2 } dy\\
&=& \color{red}{\left[ - 2 \frac{ \exp(-y^2) - 1 }{y} \right]_0^\infty}
- \color{blue}{\int_0^\infty \exp(-y^2) dy}\\
&=& \underbrace{\color{red}{- 2 \lim_{y \rightarrow 0} \frac{ \exp(-y^2) - 1 }{y}}}_{\textit{use l'Hôpital}} - \color{blue}{2 \sqrt{\pi}}\\
&=& \color{red}{4 \lim_{y \rightarrow 0} \exp(-y^2)} - \color{blue}{2 \sqrt{\pi}}\\
&=& \color{blue}{- 2 \sqrt{\pi}}.
\end{eqnarray}
$$
</s>