# How to find this integration?

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

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

• Is $\beta$ real, natural, or complex? – Axoren Feb 17 '16 at 5:27
• Seems like some variant of the $\Gamma$ function. – Henricus V. Feb 17 '16 at 5:27
• @Axoren, it is a real number – Dimitrios Feb 17 '16 at 5:28

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