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Can some please help me to find this integral.

$\int_0^\infty e^{-tx}x^{-2\beta}\text{d}x$

Is there any closed-form solution. Approximation is welcome..

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  • $\begingroup$ Is $\beta$ real, natural, or complex? $\endgroup$ – Axoren Feb 17 '16 at 5:27
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    $\begingroup$ Seems like some variant of the $\Gamma$ function. $\endgroup$ – Henricus V. Feb 17 '16 at 5:27
  • $\begingroup$ @Axoren, it is a real number $\endgroup$ – Dimitrios Feb 17 '16 at 5:28
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Let $\alpha = -2\beta + 1$. Then ($y = tx,dy = t\,dx$) $$\begin{align*} \int_0^\infty e^{-tx} x^{-2\beta} dx &= \int_0^\infty e^{-tx} x^{\alpha - 1} dx \\ &= \frac{1}{t} \int_0^\infty e^{-y} t^{-\alpha+1} y^{\alpha - 1} dy \\ &= t^{-\alpha} \int_0^\infty e^{-y} y^{\alpha - 1} dy \\ &= t^{-\alpha} \Gamma(\alpha) = t^{2\beta - 1} \Gamma(-2\beta + 1) \end{align*}$$ There are certain restrictions on $t$ and $\beta$ though.

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  • $\begingroup$ Would you please explain the restrictions in details.. $\endgroup$ – Dimitrios Feb 17 '16 at 5:55
  • $\begingroup$ @Dimitrios Obviously $t \neq 0$ for negative $2\beta - 1$. Also, $-2\beta + 1$ cannot evaluate to a nonpositive integer. $\endgroup$ – Henricus V. Feb 17 '16 at 6:27
  • $\begingroup$ The integral will only converge at zero if $\Re(\beta)<\frac{1}{2}$. $\endgroup$ – carmichael561 Feb 17 '16 at 6:46

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