MHB An integral related to beta function

AI Thread Summary
The discussion centers on proving the integral identity involving the beta function and a specific integral form. Initially, the integral was incorrectly expressed, leading to confusion, but it was later corrected to show the relationship with the beta function and Euler's integral representation. The corrected form emphasizes the transformation of the integral into a more manageable expression, facilitating the proof. Additionally, the discussion includes a series expansion that relates to hypergeometric functions, illustrating the connection to the beta function. The conversation ultimately clarifies the integral's correct formulation and its implications in mathematical analysis.
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Prove the following

$$\int^1_0 \frac{x^{r-1} (1-x)^{s-1}}{\left(ax+b(1-x)+c \right)^{r+s}}\, dx =\frac{\beta(r,s)}{ (a+c)^{r}(b+c)^{s}}$$
 
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Let $I$ denote the integral. It is obvious to see that
$$I=(b+c)^{-r-s} B(r,s) {\;}_2F_1 \left(\begin{matrix}r+s,r \\ r+s\end{matrix};\frac{b-a}{b+c} \right)\tag{1} $$
where $_2F_1(a,b;c;z)$ denotes the Hypergeometric Function. Using equation (67) of this page, we get
$$\begin{align*}
I &=(b+c)^{-r-s} B(r,s) {\;}_2F_1 \left(\begin{matrix}1,r \\ 1\end{matrix};\frac{b-a}{b+c} \right)
\\
&=B(r,s) (b+c)^{-r-s} \left(1-\frac{b-a}{b+c}\right)^{-r} \\
&=\frac{B(r,s)}{(a+c)^r(b+c)^s} \tag{2}
\end{align*}$$
 
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Shobhit said:
Let $I$ denote the integral. It is obvious to see that
$$I=(b+c)^{-r-s} B(r,s) {\;}_2F_1 \left(\begin{matrix}r+s,r \\ r+s\end{matrix};\frac{b-a}{b+c} \right)\tag{1} $$
where $_2F_1(a,b;c;z)$ denotes the Hypergeometric Function. Using equation (67) of this page, we get
$$\begin{align*}
I &=(b+c)^{-r-s} B(r,s) {\;}_2F_1 \left(\begin{matrix}1,r \\ 1\end{matrix};\frac{b-a}{b+c} \right)
\\
&=B(r,s) (b+c)^{-r-s} \left(1-\frac{b-a}{b+c}\right)^{-r} \\
&=\frac{B(r,s)}{(a+c)^r(b+c)^s} \tag{2}
\end{align*}$$
I've wanted to say this for a long time. Would you please explain the "obvious" to me?

-Dan
 
topsquark said:
I've wanted to say this for a long time. Would you please explain the "obvious" to me?

-Dan

The integral wasn't initially written correctly.

It should be

$$ \int_{0}^{1} \frac{x^{r-1} (1-x)^{s-1}}{ \Big( ax + b(1-x) + c \Big)^{s+r}} \ dx = \int_{0}^{1} x^{r-1} (1-x)^{s-1} \Big( b+c + (a-b)x \Big)^{-(r+s)} \ dx $$

$$ = (b+c)^{-(r+s)} \int_{0}^{1} x^{s-1} (1-x)^{r-1} \Big(1+ \frac{a-b}{b+c} x \Big)^{-(r+s)} \ dx $$

$$= (b+c)^{-(r+s)} \int_{0}^{1} x^{s-1} (1-x)^{r-1} \Big(1- \frac{b-a}{b+c} x \Big)^{-(r+s)} \ dx$$Then relate it to Euler's integral representation of $ {}_{2}F_{1}(a,b;c,z) $.

That is, relate it to $ \displaystyle {}_{2}F_{1}(a,b;c,z) = \frac{1}{B(b,c-b)} \int_{0}^{1} x^{b-1} (1-x)^{c-b-1} (1-tx)^{-a} \ dx$.
 
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Random Variable said:
The integral wasn't initially written correctly.

It should be

$$ \int_{0}^{1} \frac{x^{r-1} (1-x)^{s-1}}{ \Big( ax + b(1-x) + c \Big)^{s+r}} \ dx = \int_{0}^{1} x^{r-1} (1-x)^{s-1} \Big( b+c + (a-b)x \Big)^{-(r+s)} \ dx $$

Sorry guys for the confusion (Headbang) . I corrected it.
 
And without referring to equation (67),

$$ \ {}_{1}F_{2} \Big(r+s,r;r+s;\frac{b-a}{b+c} \Big) = {}_{1}F_{0} \Big(r;-;\frac{b-a}{a+c} \Big) = \sum_{n=0}^{\infty} \frac{\Gamma(r+n)}{\Gamma(r)} \Big( \frac{b-a}{b+c} \Big)^{n} \frac{1}{n!}$$

$$ = \sum_{n=0}^{\infty} \frac{(r+n-1)(r+n-2) \cdots (r)}{n!} \Big(\frac{b-a}{b+c} \Big)^{n} $$

$$ = \sum_{n=0}^{\infty} (-1)^{n} (-1)^{n} \frac{r(r+1) \cdots (r+n-1)}{n!} \Big(\frac{b-a}{b+c} \Big)^{n}$$

$$ = \sum_{n=0}^{\infty} (-1)^{n} \frac{(-r)(-r-1) \cdots (-r-n+1)}{n!} \Big( \frac{b-a}{b+c} \Big)^{n} $$

$$ = \sum_{n=0}^{\infty} (-1)^{n} \binom{-r}{n} \Big( \frac{b-a}{b+c} \Big)^{n} = \sum_{n=0}^{\infty} \binom{-r}{n}\Big( - \frac{b-a}{b+c} \Big)^{n} = \Big(1- \frac{b-a}{b+c} \Big)^{-r}$$
 
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