proof that a density proportional to Gaussian is Gaussian

Question

I try to develop bayesian estimation for one dimensional Gaussian with unknown $\mu$ and known $\sigma$. I got

$$p(x\mid D) = \int p(x\mid\mu)p(\mu\mid D) \, d\mu =\int \frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right)\frac{1}{\sigma_n\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{\mu-\mu_n}{\sigma_n}\right)^2\right)\,d\mu = \frac{1}{2\pi\sigma\sigma_n}\exp\left(-\frac{1}{2}\frac{\left(x-\mu_n\right)^2}{\sigma^2+\sigma_n^2}\right)f(\sigma,\sigma_n)$$ where $$f(\sigma,\sigma_n) = \int \exp\left(-\frac{1}{2}\frac{\sigma^2+\sigma_n^2}{\sigma^2\sigma_n^2}\left(\mu-\frac{\sigma_n^2x+\sigma^2\mu_n}{\sigma^2+\sigma_n^2}\right)^2\right)d\mu$$.

Now, It says that $p(x\mid D)$ is proportional to $$\exp\left(-\frac{1}{2}\frac{\left(x-\mu_n\right)^2}{\sigma^2+\sigma_n^2}\right)$$ as a function of x and hence $p(x\mid D)$ is normally distributed with mean $\mu_n$ and variance $\sigma^2+\sigma_n^2$.

I am asking for an explanation and/or proof for why

$p(x\mid D)$ is proportional to the $\exp$ term $\Rightarrow p(x\mid D)$ distibuted normally.

Answer 1

The condition that a probability density must integrate to 1 gives you that its normally distributed.