I've got two distributions
$$
p_1(x) = \cfrac{1}{\sqrt{2\pi}\sigma_1}\cdot e^{-x^2/2\sigma_1^2}
$$
and similarly
$$
p_2(x) = \cfrac{1}{\sqrt{2\pi}\sigma_2}\cdot e^{-x^2/2\sigma_2^2}
$$
I'm told that using the convolution theorem is the way to go so we'll start from there.
I know that $F(p_1(x)) = \int\limits_{-\infty}^\infty p_1(x) e^{2\pi ikx} dx$
so
$$
\begin{align}
F(p_1(x))F(p_2(x))
&=
\int\limits_{-\infty}^\infty \cfrac{1}{\sqrt{2\pi}\sigma_1}\cdot e^{-x^2/2\sigma_1^2} e^{2\pi ikx} dx
\cdot
\int\limits_{-\infty}^\infty \cfrac{1}{\sqrt{2\pi}\sigma_2}\cdot e^{-x^2/2\sigma_2^2} e^{2\pi ikx} dx \\
&=
\cfrac{1}{\sqrt{2\pi}\sigma_1}\cdot
\cfrac{1}{\sqrt{2\pi}\sigma_2}
\int\limits_{-\infty}^\infty e^{-x^2/2\sigma_1^2} e^{2\pi ikx} dx
\cdot
\int\limits_{-\infty}^\infty e^{-x^2/2\sigma_2^2} e^{2\pi ikx} dx \\
&=
\cfrac{1}{2\pi\sigma_1 \sigma_2}
\int\limits_{-\infty}^\infty e^{-x^2/2\sigma_1^2} e^{2\pi ikx} dx
\cdot
\int\limits_{-\infty}^\infty e^{-x^2/2\sigma_2^2} e^{2\pi ikx} dx \\
\end{align}
$$
And then I'm not sure what to do from here… I know that I'll take the inverse Fourier transform at the end and that will reveal the final gaussian distribution but I'm not sure how to evaluate these next couple steps… I've never used Fourier before and haven't taken a class that uses integrals in a complex space.
Best Answer
Let's be even more general by computing the convolution of $f_j:=\frac{1}{\sigma_j\sqrt{2\pi}}\exp\frac{-(x-\mu_j)^2}{2\sigma_j^2},\,j\in\{1,\,2\}$. First note $f_j$ has Fourier transform$$\begin{align}\int_{\Bbb R}\frac{1}{\sigma_j\sqrt{2\pi}}\exp\frac{-(x-\mu_j)^2+4\pi\sigma_j^2ikx}{2\sigma_j^2}dx&=\Bbb E[\exp2\pi ikX|X\sim N(\mu_j,\,\sigma_j^2)]\\&=\exp(2\pi ik\mu_j-2\pi^2k^2\sigma_j^2).\end{align}$$Now use$$\begin{align}f_1\ast f_2&=\mathcal{F}^{-1}(\mathcal{F}f_1\cdot\mathcal{F}f_2)\\&=\mathcal{F}^{-1}\exp(2\pi ik(\mu_1+\mu_2)-2\pi^2k^2(\sigma_1^2+\sigma_2^2))\\&=\frac{1}{\sqrt{2\pi(\sigma_1^2+\sigma_2^2)}}\exp\frac{-(x-\mu_1-\mu_2)^2}{2(\sigma_1^2+\sigma_2^2)}.\end{align}$$