Let's consider $X$ from beta distribution $B(\theta, 1)$:
$$f(x, \theta) = \theta x^{\theta - 1}\mathbb1_{(0, 1)}(x)$$
I want to derive best test's critical region when: $H_0: \theta \ge\theta_0$ and $H_1:\theta < \theta_0$.
My work so far
Let $\theta_1 \ge \theta_0$ and $\theta_2 < \theta_0$. Then we have:
$$\frac{L(\theta_1, x)}{L(\theta_2, x)} = \frac{\theta_1^n (\prod_{i=1}^nx_i)^{\theta_1}}{\theta_2^n (\prod_{i=1}^nx_i)^{\theta_2}} $$
We are looking for a $k$ such that:
$$P_{\theta_1}(\frac{L(\theta_1, x)}{L(\theta_2, x)} > k) = \alpha $$ $$P((\frac{\theta_1}{\theta_2})^n \cdot (\prod_{i=1}^n x_i)^{\theta_1 - \theta_2} > k) \Leftrightarrow P(\prod_{i=1}^n x_i > (k(\frac{\theta_2}{\theta_1})^n))^{\theta_2 - \theta_1}$$
For simplicity let's denote $k' = (k(\frac{\theta_2}{\theta_1})^n))^{\theta_2 - \theta_1}$. To recall we're looking for such $k'$ that
$$P_{\theta_1}(\prod_{i=1}^n x_i > k') = \alpha$$ $$P_{\theta_1}(\sum_{i = 1}^n \ln(x_i) > \ln(k')) = \alpha$$ $$P_{\theta_1}(-\sum_{i=1}^n \ln(x_i) > -\ln(k')) = \alpha$$
Now we can use fact that when $X_1, X_2,...,X_n \sim B(\theta, 1)$ then $-\sum_{i=1}^n \ln(X_i) \sim Gamma(n, \frac 1 \theta)$:
$$1 - F_{\text{Gamma}(n, \frac{1}{\theta_1}}(-\ln(k'))) = \alpha$$
$$-\ln k' = F_{\text{Gamma}(n, \frac{1}{\theta_1})}^{- 1}(1 - \alpha)$$
Can I ask you to whether my calculations make any sense so far?