Gamma distribution family and sufficient statistic Let X $=X_1,...X_n$ be a random sample iid from the probability density function: 
$$ f(x;\theta)=\frac{\Gamma(\theta)\sin(\pi\theta)\theta^{1-\theta}}{\pi}e^{-\theta x}x^{-\theta}$$ $x>0, 0<\theta < 1$

Is the distribution a member of the exponential family of probability
  distributions?

we know that the exponential family member distribution have the following form: 
$$ f_x(x;\theta)=c(\theta)g(x)\exp \Big\{\sum_{j=1}^{l}Q_j(\theta)T_j(x) \Big\}$$
SOLUTION:
$c(\theta)=\frac{\Gamma(\theta)\sin(\pi\theta)\theta^{1-\theta}}{\pi}$
$h(x)=1$
$\nu(\theta)=-\theta$
$T(x)=x + \log(x)$
However i am not sure how these were obtained ? I am aware that  $e^{-\theta x}x^{-\theta} = e^{-\theta(x+\log(x))}$. Then $T(x)$ contains all expression containing x and $c(\theta)$ everything containing $\theta$. What is the purpose of $h(x)$ and $\nu(\theta)?$ I am also interested in finding the sufficient statistic for parameter $\theta$. 
Now the joint density of $X_1,...,X_n$ is
$$ f(x;\theta)= c(\theta)^n e^{-\theta \sum (x_1 + \log x_i)}$$ 
which comes from the fisher factorization. 
Therefore $\sum (x_1 + \log x_i)$ is sufficient statistic for $\theta$. 
 A: Using the reflection formula $$\Gamma(\theta)\Gamma(1-\theta)=\frac{\pi}{\sin \theta\pi}\quad,0<\theta<1$$ your population density is simply
$$f_{\theta}(x)=\frac{e^{-\theta x}x^{-\theta}\theta^{1-\theta}}{\Gamma(1-\theta)}\mathbf 1_{x>0}\quad,0<\theta<1$$
(This 'simplification' is obviously not needed for the given problem)

Let $f_{\theta}(x)$, with $\theta\in\Theta$ and $x\in\mathfrak X$, be the pdf of a random variable $X$. Assume that


*

*The parameter space $\Theta$ is an open interval.

*The support $\mathfrak X=\{x:f_{\theta}(x)>0\}$ is independent of $\theta$.


We say that the joint density $f_{\theta}(\mathbf x)$ of $(X_1,\cdots,X_n)$ is a member of the one-parameter exponential family if it can be written as 
\begin{align}
f_{\theta}(\mathbf x)=\exp\left[A(\theta)T(\mathbf x)+B(\theta)+C(\mathbf x)\right]
\end{align}
, where $A(\theta)$ and $B(\theta)$ are real valued functions of $\theta$ only, and $C(\mathbf x)$ and $T(\mathbf x)$ are real valued functions of $\mathbf x$ only.
All the functions mentioned above come from  a simple analysis of the case of equality in the Cramer-Rao lower bound, from which the one-parameter exponential family can be derived.

Yes, the family of distributions $\{f_{\theta}:0<\theta<1\}$ is a member of the one-parameter exponential family (the first two assumptions are seen to be valid) since the joint density of the sample $\mathbf X=(X_1,X_2,\cdots,X_n)$ is of the form 
\begin{align}
f_{\theta}(\mathbf x)&=\prod_{i=1}^nf_{\theta}(x_i)
\\&=\left(\frac{\theta^{1-\theta}}{\Gamma(1-\theta)}\right)^n\exp\left(-\theta\sum_{i=1}^nx_i\right)\left(\prod_{i=1}^nx_i\right)^{-\theta}\mathbf1_{x_1,\cdots,x_n>0}
\\&=\exp\left[-\theta\sum_{i=1}^n(\ln x_i+x_i)+n\ln \left(\frac{\theta^{1-\theta}}{\Gamma(1-\theta)}\right)+\ln (\mathbf 1_{\min x_i>0})\right] 
\end{align}
Thus we have expressed the joint density in the general structure. That is, for some functions $A,B,C$ and $T$, we have expressed the joint density as
\begin{align}
f_{\theta}(\mathbf x)&=\exp\left[A(\theta)T(\mathbf x)+B(\theta)+C(\mathbf x)\right]
\\&=g(\theta, T(\mathbf x))h(\mathbf x)
\end{align}
, where $g(\theta,T(\mathbf x))=\exp\left[A(\theta)T(\mathbf x)+B(\theta)\right]$ depends on $\theta$ and on $x_1,\cdots,x_n$ through $T$, and $h(\mathbf x)=\exp(C(\mathbf x))$ is independent of $\theta$.
So it is justified that a sufficient statistic for $\theta$ by Factorization theorem is 
$$T(\mathbf X)=\sum_{i=1}^n(\ln X_i+X_i)$$
