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How to integrate $x^{1/2}e^{-x}$ using integration by parts?

Answer should be $\left(-\sqrt{x} e^{-x}+(1/2)\sqrt{\pi} \mbox{erf}(\sqrt{x})\right)+c$

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You can figure out what u and v were chosen from the first term of the answer, since you know what that is. That should get you closer. –  Brian Feb 22 '11 at 13:20

2 Answers 2

I am not familiar with using e.r.f but i shall tell you the procedure. Take $u= \sqrt{x}$ and $dv = e^{-x}$. So you have $du = \frac{1}{2 \cdot \sqrt{x}} \ \rm{dx}$ and $v = -e^{-x}$. Now using the integration formula we have,

\begin{align*} \int \sqrt{x} e^{-x} \ \textrm{dx} &= \Bigl[ u \cdot v ] - \int v \cdot \ \textrm{du} \\ & = \Bigl [ -\sqrt{x} \cdot e^{-x}\Bigr] + \frac{1}{2} \int \frac{e^{-x}}{\sqrt{x}} \ \textrm{dx} \end{align*}

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en.wikipedia.org/wiki/Error_function –  t.b. Feb 22 '11 at 13:18
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I just wanted to point out what $\operatorname{erf}$ means and that it is common. –  t.b. Feb 22 '11 at 13:20
    
@Theo: Ok. Theo no problem. –  anonymous Feb 22 '11 at 13:22
    
If you have following \begin{align*} \int \sqrt{x} e^{-x} \ \textrm{dx} &= \Bigl[ u \cdot v ] - \int v \cdot \ \textrm{du} \\\ & = \Bigl [ -\sqrt{x} \cdot e^{-x}\Bigr] + \frac{1}{2} \int \frac{e^{-x}}{\sqrt{x}} \ \textrm{dx} \end{align*} How you get erf(\sqrt(x)) from there without substitutions –  laovultai Feb 22 '11 at 18:29

In the comment:

How you get erf(\sqrt(x)) from there without substitutions

If you leave the $1/2$ in the integral and do the most obvious substitution (recall your first step?), you will get something that should look like the definition of the function in @t.b.'s link.

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