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

Question

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<b<\infty\\ &=\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 1

Hint:

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

Answer 2

It's the gamma function: $$\int_0^\infty x^{1/2-1}e^{-x}dx=\Gamma(1/2)=\sqrt{\pi}.$$