# Integration by parts

## 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

$$\left.\Gamma(\alpha)(a+bx)^{-\alpha}\right|_0^\infty + \int^\infty_0 \Gamma(\alpha) b\alpha(a+bx)^{-\alpha-1}dx = 0$$

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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No, it is not correct, since $\Gamma(\alpha)$ is a definite integral of that expression from $0$ to $\infty$, whereas for $u$ you need the indefinite integral of that expression. –  joriki Feb 27 '11 at 10:42

## 1 Answer

Just as an aside, this is the result from Wolfram Alpha

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I can't see the result. –  SMH Feb 27 '11 at 13:20
@SMH: I don't know why, it works fine for me. This is the imgur link: i.imgur.com/2IrTU.gif –  Simon Feb 27 '11 at 20:59
thank you. Now I can see it. –  SMH Feb 28 '11 at 19:43