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$$\int_{0}^{1} \frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}dx = \frac{B(m,n)}{(a+b)^ma^n}$$

I have to use some kind of substitution but i do not understand what i use and why ?

Thanks

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  • $\begingroup$ How do you define the Beta function? $\endgroup$ – Brevan Ellefsen Feb 10 '16 at 14:21
  • $\begingroup$ @BrevanEllefsen It is standard definition, why? $\endgroup$ – Taylor Ted Feb 10 '16 at 14:23
  • $\begingroup$ there are different ways to represent functions like this. Just making sure you don't have a specific definition you have to use, in case one of the answer uses a different one! Trust me, it happens a LOT on this site that people get annoyed an answer used a different definition -_- $\endgroup$ – Brevan Ellefsen Feb 10 '16 at 14:25
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    $\begingroup$ $$t=\frac{(a+b)x}{a+bx}$$ $\endgroup$ – Did Feb 10 '16 at 14:33
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    $\begingroup$ Did you at least try to see if it worked and how? Really zero personal input before receiving help and after receiving help... $\endgroup$ – Did Feb 10 '16 at 16:13
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Let's try the substitution

$$x=\frac{1-y}{1+cy}$$

so that $1-x=(1+c)\frac{y}{1+cy}$.

Then, when $x=0$, $y=1$ and when $x=1$, $y=0$. We also have $dx=-(1+c)\frac{1}{(1+cy)^2}\,dy$. Then, we can write

$$\begin{align}\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\int_0^1 \frac{\left(\frac{1-y}{1+cy}\right)^{m-1}\left((1+c)\frac{y}{1+cy}\right)^{n-1}}{\left(a+b\frac{1-y}{1+cy}\right)^{m+n}}\,(1+c)\frac{1}{(1+cy)^2}\,dy\\\\ &=\frac{(1+c)^n}{(a+b)^{n+m}}\int_0^1\frac{y^{n-1}(1-y)^{m-1}}{\left(1-\frac{b-ac}{a+b}y\right)^{m+n}} \end{align}$$

Choosing $c=b/a$, we obtain

$$\begin{align} \int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\frac{1}{a^n(a+b)^m}\int_0^1 y^{n-1}(1-y)^{m-1}\,dy\\\\ &=\frac{1}{a^n(a+b)^m}B(m,n) \end{align}$$

And we are done!

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  • $\begingroup$ Please tell me how do i know which substitution to use here? $\endgroup$ – Taylor Ted Feb 10 '16 at 15:00
  • $\begingroup$ @TaylorTed You may use the one in this answer. But since $B(m,n)=B(n,m)$, the one provided in the comment by Did also works. That is, if one substitutes $y=1-t$, then the two substitutions lead to the same final result. - Mark $\endgroup$ – Mark Viola Feb 10 '16 at 15:02
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    $\begingroup$ @TaylorTed I'm not sure what is meant by "these types of questions." There is no standard substitution for all problems. $\endgroup$ – Mark Viola Feb 10 '16 at 16:46
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    $\begingroup$ @Dr.MV: Do you realize what you just did ? $\endgroup$ – Lucian Feb 10 '16 at 18:37
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    $\begingroup$ @Lucian Yes, I do. The references embedded in your comment go beyond the exercise herein. - Mark $\endgroup$ – Mark Viola Feb 10 '16 at 18:47

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