Fourier transforms are frequently used to solve partial differential equations defined over infinite spatial domains. So, let us use this Fourier transform pair to solve the following heat equation $$\begin{align} u_t&=ku_{xx},\qquad -\infty<x<\infty,\\ u\left(x,0\right)&=f\left(x\right). \end{align}$$ To introduce some notation, let $\mathcal F$ and $\mathcal{F}^{-1}$ stand for the Fourier transform and inverse Fourier transform operators, respectively. It follows that $$\begin{align} \mathcal F\left(u_t\right)&=\mathcal F\left(ku_{xx}\right),\\ \frac1{2\pi}\int_{-\infty}^\infty \frac\partial{\partial t}u\left(x,t\right)e^{i\omega x}\,dx&=\frac k{2\pi}\int_{-\infty}^\infty \frac{\partial^2}{\partial x^2}u\left(x,t\right)e^{i\omega x}\,dx,\\ \frac\partial{\partial t}U\left(\omega,t\right)&=-k\omega^2U\left(\omega,t\right). \end{align}$$ This is an ordinary differential equation whose solution is $$ U\left(\omega,t\right)=c\left(\omega\right)\exp\left(-k\omega^2t\right). $$ Applying the Fourier transform to the initial condition yields $$\begin{align} \mathcal F\left[u\left(x,0\right)\right]&=\mathcal F\left[f\left(x\right)\right],\\ U(\omega,0)&=\frac1{2\pi}\int_{-\infty}^\infty f\left(x\right)e^{i\omega x}\,dx\\ &=c\left(\omega\right). \end{align}$$ To be able to apply the inverse Fourier transform to find $u$, we must know what the inverse Fourier transform of a Gaussian is, namely, $$ G(\omega)=\exp\left(-k\omega^2t\right). $$ Applying the definition yields $$\begin{align} \mathcal{F}^{-1}[G(\omega)]&=\mathcal{F}^{-1}\left[\exp\left(-k\omega^2t\right)\right],\\ g(x)&=\int_{-\infty}^\infty\exp\left(-k\omega^2t\right)\exp\left(-i\omega x\right)\,d\omega\\ &=\sqrt{\frac{\pi}{kt}}\exp\left(-\frac{x^2}{4kt}\right). \end{align}$$ We must also know that convolution is defined as follows: $$ F(\omega)G(\omega)=\frac1{2\pi}\int_{-\infty}^\infty f(\bar x)g(x-\bar x)\,d\bar x. $$ Finally, the inverse Fourier transform of $$ U\left(\omega,t\right)=\exp\left(-k\omega^2t\right)c\left(\omega\right) $$ turns out to be the following: $$ u(x,t)=\frac1{2\sqrt{\pi kt}}\int_{-\infty}^\infty\exp\left(-\frac{\bar{x}^2}{4kt}\right)f(x-\bar x)\,d\bar x. $$
Fourier Transforms
Let $f\left(x\right)$ be piecewise smooth, and
$$
\int_{-\infty}^\infty|f\left(x\right)|\,dx<\infty.
$$
The Fourier transform and inverse Fourier transform of $f$ are respectively defined as follows:
$$\begin{align}
F\left(\omega\right)&\equiv\frac1{2\pi}\int_{-\infty}^\infty f\left(x\right)e^{i\omega x}\,dx,\\
f\left(x\right)&\equiv\int_{-\infty}^\infty F\left(\omega\right)e^{-i\omega x}\,d\omega.
\end{align}$$
Together, they are called a Fourier transform pair.