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## Hi

I want to integrate this integral and ask if my work is correct or not. $\int^\infty_0 dx x^{\alpha-1} e^{-x} (a+bx)^{-\alpha}$

I want to integrate it by parts, so I have

$(a+bx)^{-\alpha} = v$

$-b\alpha(a+bx)^{-\alpha-1}dx = dv$

$x^{\alpha-1} e^{-x} dx = du$

$\Gamma(\alpha) = u$

now the integral becomes $\Gamma(\alpha)(a+bx)^{-\alpha}\vert\text{from 0 to}\infty + \int^\infty_0 \Gamma(\alpha) b\alpha(a+bx)^{-\alpha-1}dx = 0$

the problem is in integration by parts. Is it correct to put $\Gamma(\alpha) = u$. if it is not correct how can I compute this integral? please help.

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If you wonder why the question was closed, please have a look at the first two items in the faq (link at top of the page). – Harald Hanche-Olsen Feb 27 2011 at 9:44