What is the product integral of $(1+x)^{-(\theta+1)/\theta}$, if we consider that the product integral is from x=0 to x=n? It's easy to solve 1/theta, however, the second part is a little more complicated. In my opinion, the solution is $(n!)^{-(\theta+1)/\theta}$.
Sorry for my English, I know that's horrible. And also my knowledge about greek symbols.
WHICH product integral?
The geometric product integral: $\prod_a^b f(x)^{\operatorname d x} = {\sf e}^{\int_a^b \ln f(x) \operatorname d x}$
$\begin{align} \prod_{0}^{\theta} \left((1+x)^{-(\theta+1)/\theta}\right)^{\operatorname{d} x} & = {\sf e}^{-((\theta+1)/\theta)\int_0^\theta \ln(1+x) \operatorname{d}x } \\ & = {\sf e}^{-((\theta + 1)/\theta)((1+\theta)\ln(1+\theta)-\theta)} \\ & = {(1+\theta)}^{-(\theta + 1)^2/\theta}{\sf e}^{(\theta+1)} \end{align}$
The Volterra product integral: $\prod_a^b (1+f(x))\operatorname d x = {\sf e}^{\int_a^b f(x)\operatorname d x}$
$\begin{align} \prod_{0}^{\theta} \left(1+(1+x)^{-(\theta+1)/\theta}-1\right)\operatorname{d} x & = {\sf e}^{\int_0^\theta (1+x)^{-(\theta+1)/\theta}-1 \operatorname{d}x } \\ & = {\sf e}^{-\theta(1+\theta)^{-1/\theta}} \end{align}$