Integrating Gaussian white noise over a Gaussian density

Question

I have encountered the following integral: $$ \int_{- \infty}^{+ \infty} X(\theta) \frac{\exp \big \{-\frac{\theta^2}{2\sigma^2} \big \}}{\sigma\sqrt{2 \pi}}d\theta $$ where, for fixed $\theta = \theta_0$, $X(\theta_0) \sim N(0, 1)$. $X(\theta_1)$ and $X(\theta_2)$ are independent if $\theta_1 \neq \theta_2$. Hence $X(\theta)$ is Gaussian white noise indexed by $\theta$.

I am not sure about how to interpret this expression. In particular, by thinking about the integral as the limit of a sum: $$ \frac{1}{N}\sum_{i=1}^N X(\theta_i), \;\;\;\;\;\;\; X(\theta_i) \sim N(0, 1) \;\;\;\;\text{and}\;\;\;\;\; \theta_i \sim N(0, \sigma) $$ for $i = 1, \dots, N$, I would say that it should converge to zero in probability, but I'm not completely sure.

Answer 1

We have by assumption, $Y = X(\theta) \sim N(0,1)$. As corrected, the integral is, by the so-called "law of the unconscious statistician",

$$ \int_{- \infty}^{+ \infty} X(\theta) \frac{\exp \big \{-\frac{\theta^2}{2\sigma^2} \big \}}{\sigma\sqrt{2 \pi}}d\theta = \int_{- \infty}^{+ \infty} X(\theta) f_{\theta}(\theta)d\theta = E(X(\theta))=E(Y) = 0$$

DISCUSSION
Responding to @whuber 's comment, and pending clarification from the OP, I understand $X(\theta)$ as a function of $\theta$, which in turn, by the information in the question, is a $N(0,\sigma)$ random variable. Then the density of $\theta$ is $$f_{\theta}(\theta) = \frac{\exp \big \{-\frac{\theta^2}{2\sigma^2} \big \}}{\sigma\sqrt{2 \pi}}$$ and the result I give above is immediate -but I may have misunderstood the information provided in the question regarding the meaning of "indexing".