Integral Calc: Integrated by Parts - Is it Correct?

[smh]

Hi
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I want to integrate this integral and ask if my work is correct or not.
$$\int^\infty_0 dx x^{\alpha-1} e^{-x} (a+bx)^{-\alpha}$$
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I want to integrate it by parts, so I have
$$(a+bx)^{-\alpha} = v$$
$$-b\alpha(a+bx)^{-\alpha-1}dx = dv$$
$$x^{\alpha-1} e^{-x} dx = du$$
$$\Gamma(\alpha) = u$$
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now the integral becomes
$$\Gamma(\alpha)(a+bx)^{-\alpha}\vert\text{from 0 to}\infty + \int^\infty_0 \Gamma(\alpha) b\alpha(a+bx)^{-\alpha-1}dx = 0$$
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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.

 

[HallsofIvy]

No, that is not correct. Taking [itex]du= x^{\alpha-1} e^{-x} dx[/tex]
you want u to be the anti-derivative of that as a function of x. But [itex]\Gamma(\alpha)[itex] is the definite integral from 0 to infinity.