Fourier Transform with respect to x to show that the solution for the problem $$ \left\{ \begin{array}{cc} k \frac{d^2u}{dx^2}=\frac{du}{dt}, & -\infty<x<\infty, \space \space t>0,\\ u \rightarrow 0, & \text{ as } x \rightarrow \pm\infty,\\ u(x,0)=e^{-lxl}, & -\infty<x<\infty . \end{array}\right. $$
is given by $$u(x,t)=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\cos(sx)}{1+s^2}\mathrm{e}^{-ks^2t}\; ds $$
Making partial Fourier transform with respect to $x\mapsto \xi$, so that $$u(x,t)\mapsto \hat u(\xi,t)=\int_{-\infty}^\infty u(x,t) e^{-i\xi x}\mathrm dx$$ we arrive to \begin{align} \frac{\partial\hat u}{\partial t}&=−k\xi^2\hat u\tag 1\\ \hat u|_{t=0}&=\hat g(\xi)=\mathcal F\left\{e^{-|x|}\right\}=\frac{2}{1+\xi^2}\tag 2 \end{align} Indeed, $\frac{\partial}{\partial x}\mapsto i\xi$ and therefore $\frac{\partial^2}{\partial x^2}\mapsto -\xi^2$.
Note that (1) is an ODE and solving it we arrive to $\hat u=A(\xi)\,e^{−k\xi^2t}$; plugging into (2) we find that $A(\xi)=\hat g(\xi)$ and therefore $\hat u(\xi,t)=\hat g(\xi)\,e^{−k\xi^2t}$.
By inverse Fourier transform $$ u(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat g(\xi)\,e^{−k\xi^2t}e^{i\xi x}\mathrm d \xi=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{e^{−k\xi^2t}}{1+\xi^2} e^{i\xi x}\mathrm d \xi $$ Observing that the real part of the integrand is even and the imaginary part is odd on a symmetric interval, we have that the integral becomes $$ u(x,t)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\cos(\xi x)}{1+\xi^2} e^{−k\xi^2t}\mathrm d \xi $$