Show that $g_1(\mu_1,\sigma_1)\ast g_2(\mu_2,\sigma_2)=g\left(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2}\right)$ The definition (convention) I have been using for the Fourier transform is 
$$\mathscr{F}[f(t)]=g(\omega)=\frac{1}{\sqrt{2 \pi}}\int_{t=-\infty}^{\infty}f(t)e^{i\omega t}dt\tag{1}$$
and the inverse as 
$$\mathscr{F}^{-1}[g(\omega)]=f(t)=\frac{1}{\sqrt{2 \pi}}\int_{\omega=-\infty}^{\infty}g(\omega)e^{-i\omega t}d\omega$$


Let $g_1(t; \mu_1,\sigma_1)$ and $g_1(t;\mu_2,\sigma_2)$ be two Gaussian functions of $t$ with mean $\mu$ and width
  $\sigma$ as indicated. 
Show that $g_1(\mu_1,\sigma_1)\ast g_2(\mu_2,\sigma_2)=g\left(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2}\right)$

By the convolution theorem
$$\mathscr{F}(g_1 \ast g_ 2)=\sqrt{2\pi}\mathscr{F}(g_1)\mathscr{F}(g_2)\tag{2}$$ and insertion of $(1)$ into $(2)$
$\mathscr{F}(g_1 \ast g_ 2)=\sqrt{2\pi}\left[\frac{1}{\sqrt{2 \pi}}\operatorname{\Large\int}_{t_1=-\infty}^{\infty}\exp\left(-\frac{(t_1-\mu_1)^2}{2\sigma_1^2}\right)e^{i\omega t_1}dt_1\right]\left[\frac{1}{\sqrt{2 \pi}}{\Large\int}_{t_2=-\infty}^{\infty}\exp\left(-\frac{(t_2-\mu_2)^2}{2\sigma_2^2}\right)e^{i\omega t_2}dt_2\right]$
it is at this point for which I am completely stuck and don't know how to proceed to complete the proof.

The answer given by the author is: 


I don't understand the authors' solution for 3 reasons; firstly, I don't understand why there aren't any integral signs on the RHS of equation $(20)$ & $(21)$. 
Secondly, $t$ and $\omega$ are the Fourier pairs in this transformation so why is the author using $\mu_1$ and $\mu_2$ in the complex exponentials?
Lastly, the author mentions the translation property for Fourier transforms:
$$\mathscr{F}(t-t_0)=e^{i \omega t_0} g(\omega)$$ 
but I fail to see how this is being utilized in the answer.
Is there anyone that could please help me understand the authors' solution or give me any hints or tips about how to complete the proof I began?
 A: The trick is to simplify everything be completing the square and recognizing that any normal distribution will integrate to 1.
For example
$$\mathscr{F}(g_1) = \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2\pi\sigma_1^2}} 
\int_{-\infty}^{\infty}\exp(-\frac{(t_1-\mu_1)^2}{2\sigma_1^2} + i\omega t_1)dt_1$$
by completing the square
$$=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi\sigma_1^2}}
\int_{-\infty}^{\infty}\exp(-\frac{t_1^2-2t_1(\mu_1+i\sigma_1^2\omega)+(\mu_1+i\sigma_1^2\omega)^2 - 2\mu_1 i\sigma_1^2\omega + 4\sigma_1^4\omega^2}{2\sigma_1^2})dt_1
$$
$$=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi\sigma_1^2}}
\int_{-\infty}^{\infty}\exp(-\frac{(t_1-(\mu_1+i\sigma_1^2\omega))^2}{2\sigma_1^2})
\exp(\mu_1 i\omega - 2\sigma_1^2\omega^2)
dt_1
$$
Integrating out the normal distribution with mean $\mu_1+i\sigma_1^2\omega$ and variance $\sigma_1^2$ makes this
$$=\frac{1}{\sqrt{2\pi}}\exp(\mu_1 i\omega - 2\sigma_1^2\omega^2)$$
Similarly
$$\mathscr{F}(g_2) = \frac{1}{\sqrt{2\pi}} \exp(\mu_2 i\omega - 2\sigma_2^2\omega^2)$$
So 
$$\mathscr{F}(g_1 * g_2) = \frac{1}{\sqrt{2\pi}}
\exp((\mu_1+\mu_2) i\omega - 2(\sigma_1^2+\sigma_2^2)\omega^2)$$
Then the inverse Fourier transform of this is
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}
\exp((\mu_1+\mu_2-t) i\omega - 2(\sigma_1^2+\sigma_2^2)\omega^2)
d\omega
$$
$$=\frac{1}{2\pi}\int_{-\infty}^{\infty}
\exp(-\frac{\omega^2-i\omega\frac{\mu_1+\mu_2-t}{2(\sigma_1^2+\sigma_2^2)}
-(\frac{\mu_1+\mu_2-t}{2(\sigma_1^2+\sigma_2^2)})^2
+(\frac{\mu_1+\mu_2-t}{2(\sigma_1^2+\sigma_2^2)})^2}
{\frac{1}{2(\sigma_1^2+\sigma_2^2)}})
d\omega
$$
$$=\frac{1}{2\pi}\int_{-\infty}^{\infty}
\exp(-\frac{(\omega^2-i\frac{\mu_1+\mu_2-t}{2(\sigma_1^2+\sigma_2^2)})^2}
{\frac{1}{2(\sigma_1^2+\sigma_2^2)}})
\exp((\frac{\mu_1+\mu_2-t}{2(\sigma_1^2+\sigma_2^2)})^2)
d\omega
$$
Integrating out the normal distribution with mean $i\frac{\mu_1+\mu_2-t}{2(\sigma_1^2+\sigma_2^2)}$ and variance $\frac{1}{\sigma_1^2+\sigma_2^2}$ makes this
$$=\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_2^2)}}
\exp((\frac{t-(\mu_1+\mu_2)}{2(\sigma_1^2+\sigma_2^2)})^2)$$
which is $g(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2})$ as desired.
