Consider the following problem: $$\left\{\begin{array}{rcl} u_t - u_{xx} & = & f(x), \ \ \mathbb{R} \times (0,+\infty)\\ u(x,0) & = & e^{-|x|}, \ \ \mathbb{R} \end{array}\right.$$ with $$f(x) = \left\{\begin{array}{cc} e^{-x}, & x > 0\\ 0, & x \leq 0. \end{array}\right.$$
Note that $$\mathcal{F}(u_t) = \mathcal{F}(u_{xx} + f(x)) = -\xi^2\mathcal{F}(u) + \mathcal{F}(f).$$
In addition, we also have $$\mathcal{F}(f) = \dfrac{1-i\xi }{1 + \xi^2}.$$
So, we will analyze the problem: $$\left\{\begin{array}{rcl} \widehat{u_t}(\xi,t) & = & -\xi^2\widehat{u}(\xi,t) + \dfrac{1-i\xi }{1 + \xi^2}\\ \widehat{u}(\xi,0) & = & \dfrac{\sqrt{2/\pi}}{\xi^2 + 1}, \end{array}\right.$$
because $$\mathcal{F}(e^{-|x|}) = \dfrac{\sqrt{2/\pi}}{\xi^2 + 1}.$$
Now, for every $\xi \in \mathbb{R}$, we can explicitly solve the ODE, getting: $$\widehat{u}(\xi,t) = Ce^{-\xi^2t} + \dfrac{1-i\xi}{\xi^2(1 + \xi^2)},$$ where $C$ is a constant. From the initial condition, it follows that $$C = \dfrac{\sqrt{2/\pi}}{\xi^2 + 1} - \dfrac{1-i\xi}{\xi^2(1 + \xi^2)}.$$ So it follows that \begin{eqnarray} \widehat{u}(\xi,t) & = & \dfrac{\sqrt{2/\pi}}{\xi^2 + 1}e^{-\xi^2t} - \dfrac{1-i\xi}{\xi^2(1 + \xi^2)}e^{-\xi^2t} + \dfrac{1-i\xi}{\xi^2(1 + \xi^2)}\\ & = & \dfrac{\sqrt{2/\pi}}{\xi^2 + 1}e^{-\xi^2t} + \dfrac{1-i\xi}{\xi^2(1 + \xi^2)}[1-e^{-\xi^2t}]. \end{eqnarray}
Doubt: How do I return the $u$ function? I tried to calculate the inverse Fourier transform but I couldn't.