At the end of section 2 of this paper, the authors mention the integral:
$$\frac{\alpha}{Z}\int N(x_i | \mu, \sigma I) N(\mu | 0, \rho I) d \mu$$
(Note: $x_i$ and $\mu$ are $d$-dimensional.)
In the first part of section 3.1, they claim that "by a straightforward calculation," that integral evaluates to
$$\frac{\alpha}{Z} (2\pi(\rho + \sigma))^{-d/2}\exp\biggr(-\frac{1}{2(\rho + \sigma)}||x_i||^2\biggr).$$
I have tried (a few times) to obtain this result by actually evaluating the integral and completing the square by hand, but I've obtained
$$\frac{\alpha}{Z} \biggr(2\pi\frac{\sigma \rho}{\rho + \sigma} \biggr)^{d/2}\exp\biggr(-\frac{1}{2(\rho + \sigma)}||x_i||^2\biggr),$$
and I've not been able to find my mistake. I trust the paper over myself, so I must have done something incorrectly. Could someone show how to properly evaluate the integral step-by-step to obtain the paper's solution?
Let $y_i=x_i-\mu$ then $y_i \sim N(0,\sigma I)$. The distribution of $x_i$ is the same as the distribution of $y_i+\mu$ where $y_i$ and $u$ are independent. The distribution of the sum of two independent normal random variables is a normal distribution. We also have that $\sigma^2(X)+\sigma^2(Y)=\sigma^2(X+Y)$ for two independent random variables $X$, $Y$. Therefore $x_i = y_i+\mu \sim N(0,(\sigma +\rho) I)$, hence the formula. That's why it's straightforward