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I'm trying to show that

$$ \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = \frac{B(a,b)}{(1+c)^ac^b}$$

Where $$B(a,b) := \int _0^1 x^{a-1}(1-x)^{b-1}dx $$ is the "Beta function". I am supposed to use a substitution but I'm pretty much stuck. I am familiar with the basic properties of the Beta function, its relation to the gamma function etc. Any hints or advice you care to offer would be super cool.

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Let $\displaystyle \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = I$

We have ,

$(1+c)^ac^bI=\displaystyle \int _0^1 \frac{(1+c)^ac^bx^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx $

$=\displaystyle \int _0^1 \frac{(1+c)c((1+c)x)^{a-1}(c(1-x))^{b-1}}{(x+c)^{a+b}}dx $

$=\displaystyle \int _0^1 \left(\frac{(1+c)c}{(x+c)^2}\right)\left(\frac{(1+c)x}{(x+c)}\right)^{a-1}\left(\frac{c(1-x)}{(x+c)}\right)^{b-1}dx$

Let $\displaystyle y=\left(\frac{c(1-x)}{(x+c)}\right)$ and we have $\displaystyle dy=-\left(\frac{(1+c)c}{(x+c)^2}\right)dx$

Then we have the above $=-\displaystyle\int _1^0y^{b-1}(1-y)^{a-1}dy=\int_0^1y^{b-1}(1-y)^{a-1}dy=B(b,a)=B(a,b)$

So ultimately we have,

$\displaystyle(1+c)^ac^bI=B(a,b)\Rightarrow I=\frac{B(a,b)}{(1+c)^ac^b}$

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let $$\dfrac{x}{x+c}=\dfrac{1}{1+c}t$$ then $$\dfrac{c}{(x+c)^2}dx=\dfrac{1}{1+c}dt$$ then \begin{align}\int_{0}^{1}\dfrac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx&=\int_{0}^{1}\left(\dfrac{x}{x+c}\right)^{a-1}\left(\dfrac{1-x}{x+c}\right)^{b-1}\cdot\dfrac{1}{(x+c)^2}dx\\ &=\dfrac{1}{c}\dfrac{1}{1+c}\dfrac{1}{c^{b-1}}\cdot\dfrac{1}{(1+c)^{a-1}}\int_{0}^{1}t^{a-1}(1-t)^{b-1}dt\\ &=\dfrac{B(a,b)}{(1+c)^ac^b} \end{align}

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  • $\begingroup$ Thanks. Would upvote if I could $\endgroup$ – Narduar Jun 14 '13 at 8:31
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EDIT:

It appears that I deleted my answer because it was too similar to another answer in this thread. So the following is a different approach.

$ $

$$ \begin{align} \int_{0}^{1} \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}} \ dx &= \frac{1}{c^{a+b}} \int_{0}^{1} x^{a-1} (1-x)^{b-1} \left( 1+\frac{x}{c}\right) ^{-(a+b)} \ dx \\ &= \frac{B(a,b)}{c^{a+b}} {}_{2}F_{1} \left(a+b, a;a+b;-\frac{1}{c} \right) \\ &= \frac{B(a,b)}{c^{a+b}} {}_{2}F_{1} \left(a, a+b;a+b;-\frac{1}{c} \right) \end{align}$$

where I used Euler's integral representation of the hypergeometric function.

But $ \displaystyle {}_{2}F_{1} (a, b;b;z)$ is the hypergeometric representation of $(1-z)^{-a}$ (15.4.6).

Therefore,

$$ \begin{align} \int_{0}^{1} \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}} \ dx &= \frac{B(a,b)}{c^{a+b}} \left(1+\frac{1}{c} \right)^{-a}\\ &= \frac{B(a,b)}{(1+c)^{a}c^{b}} . \end{align}$$

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