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Find the values of $\alpha,\beta \in \mathbb{R}$ for which the following integral converges: $$I(\alpha,\beta) = \int_0^{\pi/2} \cos^{\alpha}(\theta)\sin^{\beta} (\theta)\ d\theta$$ and show that it is equivalent to $$B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}\ dt$$

$$$$ Attempt: $$$$ -(Values for $\beta$) $$\\$$ Since: $$ \cos^{\alpha}(\theta)\sin^{\beta} (\theta)=O_0\left(\theta^{\beta}\right)$$ We can use the Limit Comparison Test to find for which values of $\beta$ the integral converges: $$\int_0^{\pi/2}\theta^\beta\ d\theta=\left[\frac{\theta^{\beta+1}}{\beta+1}\right]_0^\frac\pi2$$ Which shows that: $$\text{If} \quad \beta+1>0\implies\beta>-1 \quad \text{the integral converges} \\ \text{If} \quad \beta\le-1 \quad \text{the integral diverges} $$ -(Values for $\alpha$) $$\\$$ Since: $$ \cos^{\alpha}(\theta)\sin^{\beta} (\theta)=O_{\pi/2}\left(\left(\frac\pi2-\theta\right)^\alpha\right)$$ We can use the Limit Comparison Test to find for which values of $\alpha$ the integral converges: $$\int_0^{\pi/2}\left(\frac\pi2-\theta\right)^\alpha\ d\theta=\left[\frac{-\left(\frac\pi2-\theta\right)^{\alpha+1}}{\alpha+1}\right]_0^\frac\pi2$$ Which shows that: $$\text{If} \quad \alpha+1>0\implies\alpha>-1 \quad \text{the integral converges} \\ \text{If} \quad \alpha\le-1 \quad \text{the integral diverges} $$ Therefore, we can conclude that the given integral converges for $\boxed{\alpha>-1,\beta>-1}$. Is this correct? I do not have the solutions to the problem. (If this is correct, are there any quicker methods?) $$$$(First approach) For the second part, I let: $$\arcsin(u)=\theta \\ \frac{du}{\sqrt{1-u^2}}=d\theta$$ which leads to: $$I(\alpha,\beta)=\int_0^1\left(1-u^2\right)^{\frac{\alpha-1}2}u^{\beta}\ du$$ Afterwards, I let: $$u^2=t \\ 2u\cdot du=dt$$ To obtain: $$I(\alpha,\beta)=\frac12\int_0^1\left(1-t\right)^{\frac{\alpha-1}2}t^{\frac{\beta-1}2}\ dt$$ which doesn't seem correct! $$$$ $$$$(Second approach) I let: $$u=\tan\left(\frac\theta2\right) \\ \frac{2du}{1+u^2}=d\theta$$ to obtain: $$I(\alpha,\beta)=\int_0^1\left(\frac{1-u^2}{1+u^2}\right)^\alpha\cdot \left(\frac{2u}{1+u^2}\right)^{\beta}\cdot\frac{2}{1+u^2}\ du$$ which doesn't lead me anywhere. Any ideas? Thank you.

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This is too long for a comment.

For the first part, you are very correct. Using Taylor expansions around $x=0$ you effectively have $$\cos ^a(x) \sin ^b(x)=x^b \left(1-\left(\frac{a}{2}+\frac{b}{6}\right) x^2+O\left(x^3\right)\right)\implies b>-1$$ and around $x=\frac\pi 2$ you have $$\cos ^a(x) \sin ^b(x)=\left(\frac{\pi }{2}-x\right)^a \left(1-\left(\frac{a}{6}+\frac{b}{2}\right) \left(x-\frac{\pi }{2}\right)^2+O\left(\left(x-\frac{\pi }{2}\right)^3\right)\right)\implies a>-1$$ all of that, as you noticed, because of the $1$ as first term inside parentheses.

For the second part,$$I(\alpha,\beta)=\frac12\int_0^1\left(1-t\right)^{\frac{\alpha-1}2}t^{\frac{\beta-1}2}\ dt$$ is very correct (and all your steps are good). It leads to $$I(\alpha,\beta)=\frac{\Gamma \left(\frac{a+1}{2}\right) \Gamma \left(\frac{b+1}{2}\right)}{2 \Gamma \left(\frac{1}{2} (a+b+2)\right)}$$ just as the original integral does.

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