integration - Calculating improper integral $int limits_{0}^{infty}frac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$

I want to calculate the improper integral $\displaystyle \int
\limits_{0}^{\infty}\dfrac{\mathrm{e}^{-x}}{\sqrt{x}}\,\mathrm{d}x$
$\DeclareMathOperator\erf{erf}$

Therefore
\begin{align}
I(b)&=\lim\limits_{b\to0}\left(\displaystyle \int \limits_{b}^{\infty}\dfrac{\mathrm{e}^{-x}}{\sqrt{x}}\,\mathrm{d}x\right) \qquad \forall b\in\mathbb{R}:0
&=\lim\limits_{b\to0}\left(\sqrt{\pi} \erf(\sqrt{b}) \right)=\sqrt{\pi}\erf(\sqrt{0})=\sqrt{\pi}
\end{align}

This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$\displaystyle \int\limits_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{1}{2}\sqrt{\pi}$$

Answer

Hint:

Just substitute $x= u^2$. So, you get
$$\int \limits_{0}^{\infty}\dfrac{\mathrm{e}^{-x}}{\sqrt{x}}\,\mathrm{d}x =2\int_0^{\infty}e^{-u^2}du$$