So, I tried to compute the expected value of continous random variable X using an integral and the typical formula given here:
$\mathbb{E}$(X)=$\int_{-\infty}^{\infty} x*f(x)) dx$
And I plugged a function in there:
$\mathbb{E}$(X)=$\int_{1}^{\infty} x*ln (\frac{\theta}{x^{\theta+1}}) dx$
with domain ranges $\theta=[2,\infty]$ and $x=[0,\infty]$
I got
$[\frac{1}{2}x^2* ln(\theta)-(\theta+1)\frac{1}{4}*x^2*(2*ln(x)-1)]_{1}^{\infty} $
Is it possible to evaluate it even further? It is far too large and not very elegant to look at if I want to compute the variance starting with this, I will have a very messy and convoluted calculation method.
So let's take a look. I think you want the integral to be evaluated from $1$ to $\infty$, in which case the answer is not a function of $x$, it is a function of $\theta$
$$I(\theta)=I=\int_1^\infty x\ln\left(\frac{\theta}{x^{\theta+1}}\right)dx=\int_1^\infty x(\ln(\theta)-(\theta+1)\ln(x))dx=\int_1^\infty \ln(\theta)x-x(\theta+1)\ln(x)dx$$
We split these apart to get:
$$\ln(\theta)\int_1^\infty xdx - (\theta+1)\int_1^\infty x\ln(x)dx=\left[\frac12\ln(\theta)x^2-\frac12x^2(\theta+1)(\ln(x)-\frac12)\right]_1^\infty=\left[\frac12x^2\left(\ln(\theta)-(\theta+1)(\ln(x)-\frac12)\right)\right]_1^\infty=\left(\frac12\cdot\left(\ln(\theta)+\frac12(\theta+1)\right)\right)-\lim_{x\rightarrow\infty}\frac12x^2\left(\ln(\theta)-(\theta+1)(\ln(x)-\frac12)\right)=$$
I could evaluate the limit here, but it should be clear that for all values of $\theta$, the limit diverges. Since we have a finite number from which infinity is being subtracted, this integral diverges.