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

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  • $\begingroup$ I corrected that, thank you! $\endgroup$ – Doesbaddel Jan 24 '19 at 11:36
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Hint:

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

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  • $\begingroup$ 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? $\endgroup$ – Doesbaddel Jan 24 '19 at 11:44
  • $\begingroup$ Yes, it is the famous Gaussian integral. $\endgroup$ – Larry Jan 24 '19 at 11:46
  • $\begingroup$ @Doesbaddel : So ist es. :-) $\endgroup$ – trancelocation Jan 24 '19 at 11:46
  • $\begingroup$ You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function $\endgroup$ – Yauhen Mardan Jan 24 '19 at 11:47
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    $\begingroup$ @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-) $\endgroup$ – trancelocation Jan 24 '19 at 11:49
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It's Gamma function: $\int_0^\infty x^{1/2-1}e^{-x}dx=Г(1/2)=\sqrt{\pi}$

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    $\begingroup$ 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. $\endgroup$ – J.G. Jan 24 '19 at 12:32

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