# 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

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?: $$\displaystyle \int\limits_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{1}{2}\sqrt{\pi}$$

• I corrected that, thank you! – Doesbaddel Jan 24 '19 at 11:36

Hint:

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

• Ok, so $\int \limits_{0}^{\infty}\dfrac{\mathrm{e}^{-x}}{\sqrt{x}}\,\mathrm{d}x =2\int_0^{\infty}e^{-u^2}du =2\cdot \frac{1}{2}\sqrt{\pi}=\sqrt{\pi}$ does the job? – Doesbaddel Jan 24 '19 at 11:44
• Yes, it is the famous Gaussian integral. – Larry Jan 24 '19 at 11:46
• @Doesbaddel : So ist es. :-) – trancelocation Jan 24 '19 at 11:46
• You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function – Yauhen Mardan Jan 24 '19 at 11:47
• @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-) – trancelocation Jan 24 '19 at 11:49

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

• In particular $\Gamma^2(1/2)=\operatorname{B}(1/2,\,1/2)=2\int_0^{\pi/2}dx=\pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral. – J.G. Jan 24 '19 at 12:32