Finding the power function of the given test. We have a density $X$ defined as $f(x,\theta)=\theta x^{\theta -1}I_{(0,1)}(x)$.
The hypothesis to test is given as follows:
$H_0:\theta \leq1$ Vs $H_1:\theta >1$
A sample size of two is selected, and the critical region is defined as follows:
$C=\{(x_1,x_2):\frac{3}{4x_1}\leq x_2\}$
I tried writing the power function, which for a general $\theta$ as follows:
$P(\frac{3}{4X_1}\leq X_2)=P(X_1X_2\geq\frac{3}{4} )$
Now since $X_1,X_2$ both are randoms we have to fix one of them, hence:
$P(X_1X_2\geq\frac{3}{4})=P(X_1\geq\frac{3}{4X_2})=\int_0^1 P(X_1\geq\frac{3}{4X_2}|X_2=x_2)f_{X_2}(x_2)dx_2$
Am I going in the right direction ? 
Evaluating this, gives a $log$ term and hence the value of the integral is coming out to be $\infty$.
What am I doing wrong ?
 A: The power function is
$$\begin{align}\pi(\theta)&=\mathbb{P}_\theta(C)\\
&=\mathbb{P}_\theta\left(X_1X_2\ge\frac{3}{4}\right)\\
&=1-\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right).\\
\end{align}$$
Now
$$\begin{align}\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right)&=\int_0^1{\mathbb{P}_\theta\left(X_1<\frac{3}{4x_2}\,\Bigg|\, X_2=x_2\right)\,f_\theta(x_2)\,dx_2}\\
&=\int_0^1\mathbb{P}_\theta\left(X_1<\frac{3}{4x_2}\right)\,f_\theta(x_2)\,dx_2\ \ \text{ (because }X_1,X_2\text{ are independent})\\
&=\int_0^1F_\theta\left(\frac{3}{4x_2}\right)\,f_\theta(x_2)\,dx_2
\end{align}$$
where $F_\theta$ is the CDF of $X$, found by integrating the density function $$f_\theta(x) = \theta x^{\theta-1}[0<x<1],$$ 
i.e.,
$$\begin{align}F_\theta(x)&=\mathbb{P}_\theta(X\le x)\\ \\
&= \int_{-\infty}^x f_\theta(x')\,dx'\\ \\
&=
\begin{cases} 
0,  & \text{if }x\le 0 \\
x^\theta, & \text{if } 0<x<1\\
1,  & \text{if }x\ge 1
\end{cases}\\ \\
&=x^\theta[0<x<1]+1[x\ge 1].\end{align}  $$
(For convenience, I'm using Iverson brackets for the indicator functions.)
Thus,
$$\begin{align}\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right)&=\int_0^1F_\theta\left(\frac{3}{4x}\right)\,f_\theta(x)\,dx\\
&=\int_0^1\left\{ \left(\frac{3}{4x}\right)^\theta\left[0<\frac{3}{4x}<1\right]\ \ +\ \ 1\left[\frac{3}{4x}\ge 1\right]\right\}\,\theta x^{\theta-1}\bigg[0<x<1\bigg]\, dx\\
&=\theta\left(\frac{3}{4}\right)^\theta\int_0^1 x^{-1}\left[0<\frac{3}{4x}<1\right]\bigg[0<x<1\bigg]\,dx\ \ \\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +\ \ \theta \int_0^1x^{\theta-1}\left[\frac{3}{4x}\ge 1\right]\bigg[0<x<1\bigg]\,\, dx\\
&=\theta\left(\frac{3}{4}\right)^\theta\int_{\frac{3}{4}}^1 x^{-1}dx  \ + \ \theta\int_0^{\frac{3}{4}}x^{\theta-1}dx\\
&=\theta\left(\frac{3}{4}\right)^\theta\log\frac{4}{3}\ + \ \left(\frac{3}{4}\right)^\theta\\
&=\left(\frac{3}{4}\right)^\theta(\theta\log\frac{4}{3}+1).
\end{align}$$
So the power function is
$$\pi(\theta) = 1 - \left(\frac{3}{4}\right)^\theta(\theta\log\frac{4}{3}+1).
$$ 
Here's a plot:

