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I'm searching a closed form for the integral of the form :

$$ \int_{0}^{\infty}e^{-ax}e^{-x^\frac{\alpha}{2}}dx $$

especially for $\alpha=8$

After several attempts, I find that the general form can be written on the basis of hypergeometric functions. what do you think ?

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    $\begingroup$ I see you wrote down "Especially for $\alpha=8$". Which one would you like us to write an answer for? $$\int_{0}^{\infty}e^{-ax}e^{-x^{4}}dx$$ $$\int_{0}^{\infty}e^{-ax}e^{-x^\frac{\alpha}{2}}dx$$ $\endgroup$
    – projectilemotion
    Commented May 2, 2017 at 12:26
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    $\begingroup$ expand $ e^{-x^4})$ into a power series and uses the integral of the gamma function $ \int_{0}^{\infty}x^{n}exp(-ax) = \Gamma (n+1)/a^{n+1} $ $\endgroup$
    – Jose Garcia
    Commented May 2, 2017 at 12:27
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    $\begingroup$ Thank you for your feedback. I need a closed form for the general case no matter the value of alpha , $\endgroup$
    – yassine hmamouche
    Commented May 2, 2017 at 12:36
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    $\begingroup$ @yassinehmamouche: a hypergeometric series is considered as a closed form? You cannot get something better than that for a generic $\alpha$. $\endgroup$
    – Jack D'Aurizio
    Commented May 2, 2017 at 12:39
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    $\begingroup$ The case $\alpha=6$ is related with Airy functions, for instance. $\endgroup$
    – Jack D'Aurizio
    Commented May 2, 2017 at 12:43

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