Statistics Question of unbiased estimator 
Could anyone give me a hint? or way to solve?
 A: To show that 
$$\frac{1}{n}\sum_{i=1}^n \log(X_i)$$
is an unbiased estimator of $\dfrac{1}{\theta}$, we need to show that 
$$E\left(\frac{1}{n}\sum_{i=1}^n \log(X_i)\right)=\frac{1}{\theta}.$$
By the linearity of expectation, the left side is equal to $E(\log(X))$. So that's what we need to calculate. By the usual formula for expectation, we need to find
$$\int_1^\infty (\log x)\theta x^{-\theta-1}\,dx.$$
To evaluate the integral, we could proceed directly to integration by parts. But it is better to make the preliminary substitution $\log x=u$. Then $\frac{1}{x}\,dx=du$. We have $x=e^u$, and therefore $dx=e^u\,du$. Substituting, we find that our integral is
$$\int_0^\infty u\theta e^{-u\theta -u}e^u\,du=\int_0^\infty u\theta e^{-\theta u}\,du.$$
The integral on the right is a stndard integration by parts. But as a probabilist, you will also recognize it as the expression for the mean of an exponential random variable with parameter $\theta$.
So either do the integration by parts, or quote the result that the exponential with parameter $\lambda$ has mean $\dfrac{1}{\lambda}$.  Either way, you will find that our integral is $\dfrac{1}{\theta}$, which is what we needed to prove to show unbiasedness of our estimator.
