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.

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: $$u(x,t)=\frac1{2\sqrt{\pi kt}}\int_{-\infty}^\infty\exp\left(-\frac{\bar{x}^2}{4kt}\right)f(x-\bar x)\,d\bar x.$$