Show that $$\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$$
My book wrote that
$$=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{4})} \cdot \frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{2})}{2\Gamma(3/4)}$$
My work says that there wasn't $\Gamma(5/4)$ in the denominator of the left one.
$$\beta(\frac{\frac{1}{2}+1}{2},\frac{1}{2})\beta(\frac{-\frac{1}{2}+1}{2},\frac{1}{2})$$
$$\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{3}{4}+1)}\cdot \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{1}{4}+\frac{1}{2})}$$
$$\frac{\Gamma(\frac{3}{4})\pi}{2\Gamma(\frac{7}{4})} \cdot \frac{\Gamma(\frac{1}{4})}{2\Gamma(\frac{3}{4})}$$
For that reason, my work doesn't match with their even answer.
Best Answer
Your method was correct, you just made a small mistake in the simplification. It is true that \begin{align} \int_0^{\dfrac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\dfrac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta \end{align} equals
\begin{align} & \beta(\frac{3}{4}, \frac{1}{2})\; \beta(\frac{1}{4}, \frac{1}{2}) \\ \\ &=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{3}{4}+\color{red}{\frac{1}{2}})}\cdot \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{1}{4}+\frac{1}{2})} \\ \\ &=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{4})} \cdot \frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{2})}{2\Gamma(3/4)} \end{align}
which is the step from your book. Then using $\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}) = \Gamma(\frac{1}{2})\Gamma(1-\frac{1}{2}) = \frac{\pi}{sin(\pi / 2)} = \pi$ is correct, and will lead to the answer.