Proving completeness of highest-order statistic using Leibnitz' Rule

Question

Suppose that $X_1,...,X_n$ are iid with common pdf given by $$f(x;\theta)=2e^{2x}\theta^{-2}I( x<log(\theta)).$$ I am tasked with finding a complete-sufficient statistic for $\theta$, and I have already shown via Neyman-factorization that $X_{(n)}$ is sufficient.

To demonstrate completeness, let $h(x_{(n)})$ be a real-valued function of $x_{(n)}$ such that $E[h(x_{(n)})]=0$. We can now express this as the following integral: $$0=\int_{-\infty}^{log(\theta)}h(x_{(n)})2n\theta^{-2n}e^{2nx_{(n)}}dx_{(n)},$$

where the expression being multiplied by $h$ is the pdf of the n-th order statistic. Now, this is where I am a little confused. First, note that because we are integrating w.r.t. the n-th order statistic, we can divide both sides by all expressions in the integral which do not contain the n-th order statistic. We now have the following: $$0=\int_{-\infty}^{log(\theta)}h(x_{(n)})e^{2nx_{(n)}}dx_{(n)}.$$

I am now inclined to use Leibnitz' rule, but in the last expression of Leibnitz' rule, where I take the partial w.r.t. $\theta$ of everything in the integral, it will be free of $h(x_{(n)})$ which is a problem, as I need to show that $h(x_{(n)})=0$.

My question is, should I not have eliminated the seemingly ancillary expressions inside the integral? Or am I making some other type of mistake?

Answer 1

The derivative of $$\int_{-\infty}^{\log(\theta)}h(x)e^{2nx}\text{d}x$$ which is the composition of the two functions $$\theta \mapsto \log(\theta)\qquad\text{ and }\qquad \xi \mapsto \int_{-\infty}^{\xi}h(x)e^{2nx}\text{d}x$$ wrt $\theta$ is $$\frac{\text{d}}{\text{d}\theta}\int_{-\infty}^{\log(\theta)}h(x)e^{2nx}\text{d}x=\frac{1}{\theta}\,h(\log(\theta))e^{2n\log(\theta)}=\theta^{2n-1}h(\log(\theta))$$ which is uniformly null iff $h\equiv 0$. There is no reason for resorting to Leibniz's rule in that case.