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Prove that:

$\int_{-1}^2 (1+x)^{p-1}(1-x)^{q-1} dx$ = $2^{p+q-1}\beta{(p,q)}.$

I tried converting the integral into the standard forms of beta function:

$\beta{(p,q)}=\int_{0}^1 (x)^{p-1}(1-x)^{q-1} dx$

$\beta(p, q) = \int_{0}^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$

using various substitutions like ${(1+x)}=t$, ${(1-x)}=t$, $(1-x)/(1+x)=t$ but they all failed eventually .

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  • $\begingroup$ Are you sure that you copied that right? I checked for some $p,q$ values and I don't think it matches. $\endgroup$ – Zacky Feb 28 at 14:44
  • $\begingroup$ Yes, I copied the question as it is printed on the book . $\endgroup$ – Alphanerd Feb 28 at 14:46
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    $\begingroup$ @Zacky. You are right. It does not match. With $x=2y-1$, the bounds become $0$ and $\frac 12$. $\endgroup$ – Claude Leibovici Feb 28 at 14:53
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    $\begingroup$ More than likely, one more typo in a textbook ! $\endgroup$ – Claude Leibovici Feb 28 at 15:12
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Hint

Focus on the antiderivative first and change variable $x=2y-1$.

This gives $$\int (1+x)^{p-1}(1-x)^{q-1}\, dx=2^{p+q-1}\int y^{p-1} (1-y)^{q-1}\,dy$$

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  • $\begingroup$ Plugging in $x=2$ in the equation $x=2y-1$ , we get $y=3/2$. However if the upper limit in the original integral is $1$, then with change of variable we get the correct form of beta function. Thank you for answering. $\endgroup$ – Alphanerd Feb 28 at 14:53

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