Heat Equation is
$$Au_{xx} = u_{t},$$ for $-\infty<x<\infty$ and $t>0$. Also, $u(x,0) \rightarrow 0$ as $ x \rightarrow \pm \infty$.
The initial condition:
$$ u(x,0) = \exp{(-|x|)},$$
for $-\infty<x<\infty$.
Want the solution form: (Using the Fourier Transform)
$$u(x,t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{cos(wx)}{1+w^2} \exp{(-Aw^2t) dw}.$$
The fundamental FT equations are:
$$F(s) = \mathcal{F[f(x)]} = \int_{-\infty}^{\infty} f(x) \exp{(-iwx)}dx$$ $$f(x) = \mathcal{F^{-1}[f(x)]} = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(s) \exp{(iwx)}dw$$
Attempt:
Taking fourier transform of both sides of the PDE,
$$-Aw^2U(w,t) = \frac{\partial U}{\partial t} (w,t)$$ Then by solving this as an ODE and using the integrating factor, $$U(w,t) = U(w,0)\exp{(-Aw^2t)}$$
Taking inverse fourier transform of both sides,
$$u(x,t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{1}{1+w^2} \exp{(iwx)} \exp{(-Aw^2t)} dw.$$
But clearly this is not the required answer. Euler's formula can be used to get
$$u(x,t) = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{cos(wx)}{1+w^2}\exp{(-Aw^2t)} dw + \frac{i}{\pi}\int_{-\infty}^{\infty} \frac{sin(wx)}{1+w^2} \exp{(-Aw^2t)} dw.$$
However, that means that the integral on the right must be equal to zero (which computationally it indeed is). But lack of knowledge about that type of integral makes this impossible. Feel as if there's something easy that is being missed. Any help would be appreciated.