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An integral related to beta function

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove the following

\(\displaystyle \int^1_0 \frac{x^{r-1} (1-x)^{s-1}}{\left(ax+b(1-x)+c \right)^{r+s}}\, dx =\frac{\beta(r,s)}{ (a+c)^{r}(b+c)^{s}}\)
 
Last edited:

Shobhit

Member
Nov 12, 2013
23

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
$$\begin{align*}
I &=(b+c)^{-r-s} B(r,s) {\;}_2F_1 \left(\begin{matrix}1,r \\ 1\end{matrix};\frac{b-a}{b+c} \right)
\\
&=B(r,s) (b+c)^{-r-s} \left(1-\frac{b-a}{b+c}\right)^{-r} \\
&=\frac{B(r,s)}{(a+c)^r(b+c)^s} \tag{2}
\end{align*}$$

 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,135

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
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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$.
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
And without referring to equation (67),

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

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

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

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

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