Under a weak prior $\pi (\theta) \propto 1$ the posterior is $\pi (\theta | x ) \propto \mathcal{L} (\theta)$ and then the normalising constant is just $\int \mathcal{L}(\theta) d\theta$. Hence, the posterior expectation is \begin{align*} E(\theta|x) & = \int \theta \pi(\theta | x) d\theta \\ & = \int \frac{\theta \mathcal{L}(\theta)}{\int \mathcal{L}(\theta) d\theta} d\theta \\ & = \frac{\int \theta \mathcal{L}(\theta) d\theta}{\int \mathcal{L}(\theta) d\theta} \end{align*}

So $\hat{\theta}$ is can be interpreted as the posterior expectation of $\theta$ under the improper prior $\pi(\theta) \propto 1$. Wikipedia tells me this is the ''generalised Bayes estimator''. Unfortunately, I don't know much about the generalised bayes estimator but know that we know it's a thing then someone else on stackexchange can help us find out more about it!

Under a weak improper prior $\pi (\theta) \propto 1$ the posterior is $\pi (\theta | x ) \propto \mathcal{L} (\theta)$ and then the normalising constant is just $\int \mathcal{L}(\theta) d\theta$. Hence, the posterior expectation is \begin{align*} E(\theta|x) & = \int \theta \pi(\theta | x) d\theta \\ & = \int \frac{\theta \mathcal{L}(\theta)}{\int \mathcal{L}(\theta) d\theta} d\theta \\ & = \frac{\int \theta \mathcal{L}(\theta) d\theta}{\int \mathcal{L}(\theta) d\theta} \end{align*}

So $\hat{\theta}$ is can be interpreted as the posterior expectation of $\theta$ under the improper prior $\pi(\theta) \propto 1$. Wikipedia tells me this is the ''generalised Bayes estimator''. Unfortunately, I don't know much about the generalised bayes estimator but know that we know it's a thing then someone else on stackexchange can help us find out more about it!