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Can someone help me to evaluate the integral $$ \int_{b}^\infty \sqrt{x} e^{-ax}dx. $$ I guessed that it has no elementary anti-derivative, and indeed substituting $x=t^2$ and then applying integration by parts resulted in $\int e^{-at^2}$.

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You can get an answer in terms of the incomplete gamma function. –  Mhenni Benghorbal Feb 20 '13 at 18:31
    
@Mhenni I know of the gamma function, but the incomplete one? I'll check it in wikipedia. I'm looking for something `simple' , if exists. –  Lior B-S Feb 20 '13 at 18:33
    
It is known that there is no expression for $\int e^{-at^2}dt$ in terms of elementary functions, I believe. –  Thomas Andrews Feb 20 '13 at 18:34
    
I already added a link. –  Mhenni Benghorbal Feb 20 '13 at 18:34
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If you put it as an answer I'll accept it. –  Lior B-S Feb 20 '13 at 18:37

2 Answers 2

up vote 4 down vote accepted

$$\int_{b}^\infty \sqrt{x} e^{-ax}dx= \frac{1}{a^{3/2}}\int_{a b}^\infty y^{1/2} e^{-y}dy = \Gamma(3/2,ab). $$

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You can write it in terms of the error function (which is an antiderivative of $\frac{2}{\sqrt{\pi}} {\rm e}^{-x^2}$) as $$ \frac {\sqrt {b}}{a}{\rm e}^{-ab} - \frac{\sqrt{\pi}}{2 a^{3/2}} \text{erf}(\sqrt{ab}) + \frac{\sqrt{\pi}}{2 a^{3/2}}$$

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Thank you for your answer. As you can see in my question, I managed to express it with the erf function. –  Lior B-S Feb 20 '13 at 18:54

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