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

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 .

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

Focus on the antiderivative first and change variable $$x=2y-1$$.
This gives $$\int (1+x)^{p-1}(1-x)^{q-1}\, dx=2^{p+q-1}\int y^{p-1} (1-y)^{q-1}\,dy$$
• 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. – Alphanerd Feb 28 at 14:53