Integral Calc: Integrated by Parts - Is it Correct?

[smh]

Hi
--

I want to integrate this integral and ask if my work is correct or not.
<br /> \int^\infty_0 dx x^{\alpha-1} e^{-x} (a+bx)^{-\alpha}<br />
----------
I want to integrate it by parts, so I have
<br /> (a+bx)^{-\alpha} = v <br />
<br /> -b\alpha(a+bx)^{-\alpha-1}dx = dv<br />
<br /> x^{\alpha-1} e^{-x} dx = du <br />
<br /> \Gamma(\alpha) = u<br />
----------
now the integral becomes
<br /> \Gamma(\alpha)(a+bx)^{-\alpha}\vert\text{from 0 to}\infty + \int^\infty_0 \Gamma(\alpha) b\alpha(a+bx)^{-\alpha-1}dx = 0<br />
----------
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 du= x^{\alpha-1} e^{-x} dx[/tex]<br /> you want u to be the <b>anti-derivative</b> of that as a function of x. But \Gamma(\alpha)is the &amp;lt;b&amp;gt;definite&amp;lt;/b&amp;gt; integral from 0 to infinity.