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 $\int_0^1 f(x)dx$ ?
 A: Since $\int_{0}^{1}\frac{1}{x^{1/2}}$ converges and $\displaystyle\lim_{x\to 0^{+}}\frac{\frac{1-e^{-x}}{x^{3/2}}}{\frac{1}{x^{1/2}}}=\lim_{x\to 0^{+}}\frac{1-e^{-x}}{x}=\lim_{x\to 0^{+}}\frac{e^{-x}}{1}=1$, 
$\;\;\;\;\;\int_{0}^{1}\frac{1-e^{-x}}{x^{3/2}}$ converges.
A: This integral is not too difficult to just go ahead and solve completely. Starting by integrating by parts,
$$\begin{align}
\int_{0}^{\infty}\frac{e^{-x}-1}{x^{3/2}}\mathrm{d}x
&=\left[-\frac{2}{x^{1/2}}(e^{-x}-1)\right]_{0}^{\infty}-\int_{0}^{\infty}\frac{2\,e^{-x}}{x^{1/2}}\mathrm{d}x\\
&=0-0-2\int_{0}^{\infty}x^{\frac12-1}e^{-x}\mathrm{d}x\\
&=-2\,\Gamma{\left(\frac12\right)}\\
&=-2\,\sqrt{\pi},
\end{align}$$
where we've substituted the definition of the Gamma function and used the fact that $\Gamma{\left(\frac12\right)}=\sqrt{\pi}$. Since the integral converges to the limit $-2\,\sqrt{\pi}$, the integral obviously converges.

Addendum:
Suppose you didn't know the value of $\Gamma{\left(\frac12\right)}$. An easy way to establish that it converges is to make a substitution that transforms the it into the Gaussian integral, whose convergence is probably common knowledge even for most beginning students. Substituting $x=u^2$, we have $u=x^{1/2}$ and $\mathrm{d}u=\frac{\mathrm{d}x}{2\,x^{1/2}}$, and hence:
$$\begin{align}
\Gamma{\left(\frac12\right)}
&=\int_{0}^{\infty}x^{\frac12-1}e^{-x}\mathrm{d}x\\
&=2\int_{0}^{\infty}e^{-x}\frac{\mathrm{d}x}{2\,x^{\frac12}}\\
&=2\int_{0}^{\infty}e^{-u^2}\mathrm{d}u.
\end{align}$$
A: Given
$$
\int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx.
$$
Use substitution $x = y^2$, then we obtain
$$
\begin{eqnarray}
\int_0^\infty \frac{ \exp(-x) - 1 }{ \sqrt{x}^3 } dx &=& 2 \int_0^\infty \frac{ \exp(-y^2) - 1 }{ y^2 } dy\\
&=& \color{red}{\left[ - 2 \frac{ \exp(-y^2) - 1 }{y} \right]_0^\infty}
- \color{blue}{\int_0^\infty \exp(-y^2) dy}\\
&=& \underbrace{\color{red}{- 2 \lim_{y \rightarrow 0} \frac{ \exp(-y^2) - 1 }{y}}}_{\textit{use l'Hôpital}} - \color{blue}{2 \sqrt{\pi}}\\
&=& \color{red}{4 \lim_{y \rightarrow 0} \exp(-y^2)} - \color{blue}{2 \sqrt{\pi}}\\
&=& \color{blue}{- 2 \sqrt{\pi}}.
\end{eqnarray}
$$
