1. ## Integral

does anyone know how to evaluate this integral

$\Phi(X) = C \cdot (b-X)^{\alpha-1}(X-a)^{\beta-1} dX$

where $C, a,b,\alpha$ and $\beta$ are constants

2. Originally Posted by chogo
does anyone know how to evaluate this integral

$\Phi(X) = C \cdot (b-X)^{\alpha-1}(X-a)^{\beta-1} dX$

where $C, a,b,\alpha$ and $\beta$ are constants
You fail to mention the limits of integration.
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This is what I will assume:
1) $b>a$
2)The limits of integration are from $x=a \mbox{ to }x=b$.
3)And $\alpha,\beta >0$.
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We can ignore the constant $C$ and instead consider:
$\int_a^b (b-x)^{\alpha - 1}(x-a)^{\beta - 1} dx$
Use the functional substitution $s=x-a$:

$\int_0^{b-a} (b-a -s)^{\alpha - 1}\cdot s^{\beta -1} ds = (b-a)^{\alpha - 1} \int_0^{b-a} \left( 1 - \frac{s}{b-a} \right)^{\alpha - 1}\cdot s^{\beta - 1} ds$

Now use the substitution $t=\frac{s}{b-a}$:

$(b-a)(b-a)^{\alpha - 1}\int_0^1 (1-t)^{\alpha - 1}\cdot t^{\beta - 1} \cdot (b-a)^{\beta - 1} dt = (b-a)^{\alpha + \beta - 1} \int_0^1 (1-t)^{\alpha - 1} t^{\beta-1} dt$

Realize that this is the Beta Integral.

Thus,
$(b-a)^{ \alpha + \beta - 1} \cdot \bold{B} (\alpha , \beta )$

We can convert this to the Gamma function.

$\boxed{ (b-a)^{\alpha + \beta - 1} \cdot \frac{\Gamma (\alpha )\cdot \Gamma (\beta )}{\Gamma (\alpha + \beta) }}$

So if $\alpha,\beta$ are positive integers we end up with a nice looking result in the end.

This is Mine 6th Post!!!

3. thanks alot. I stuck it into the wolfram integrator and got a formula using the hypergeometric funtion. This makes more sense. The limits are however

$x=t, x=s$

where $s>t$

thanks and congrats on the 6000 post