the problem : \begin{cases} u_{t} -a^2u_{xx} = f(x,t) \; & \text{in} \; \mathbb{R_+} \times(0,+\infty) \\ u = 0\; & \text{on} \; \mathbb{R_+} \times\{t = 0\} \\ u = 0, \, u_x = 0\; & \text{on} \; \{x \to +\infty \} \times(0,+\infty) \\ \end{cases}
I'm using the following conventions :
$$\hat{u}(\lambda,t) = \frac{1}{\pi}\int_{\mathbb{R_+}} u(x,t)\cos (\lambda x) \, dx$$ $$u(x,t) = 2\int_{\mathbb{R_+}} \hat{u}(\lambda,t)\cos (\lambda x) \, d \lambda$$
using an IBP and boundary conditions the problem transforms to : $$ \begin{align} \frac{1}{\pi}\int_{\mathbb{R_+}} [u_t(x,t)-a^2u_{xx}(x,t)]\cos (\lambda x) \, dx & = \frac{1}{\pi}\int_{\mathbb{R_+}} f(x,t)\cos (\lambda x) \, dx \\ \hat{u}_t(\lambda,t) -\frac{a^2}{\pi}\int_{\mathbb{R_+}} [u_{xx}(x,t)]\cos (\lambda x) \, dx & = \hat{f}(\lambda,t) \\ \hat{u}_t(\lambda,t) - \frac{a^2}{\pi}(u_x \cos(\lambda x))|_{x = 0}^{x \to \infty}+\frac{a^2\lambda}{\pi} \int_{\mathbb{R_+}} u_{x}(x,t)\sin (\lambda x) \, dx & = \hat{f}(\lambda,t) \\ \hat{u}_t(\lambda,t) + \frac{a^2\lambda}{\pi}(u \sin(\lambda x))|_{x = 0}^{x \to \infty}-\frac{a^2\lambda^2}{\pi}\int_{\mathbb{R_+}} u(x,t)\cos (\lambda x) \, dx & = \hat{f}(\lambda,t) \\ \hat{u}_t(\lambda,t) -a^2\lambda^2\hat{u}(\lambda,t) & = \hat{f}(\lambda,t) \\ \end{align}$$
multiplying both sides by $e^{-a^2\lambda^2 t}$ gives : $$ \begin{align} \hat{u}_t(\lambda,t)e^{-a^2\lambda^2 t} -a^2\lambda^2\hat{u}(\lambda,t)e^{-a^2\lambda^2 t} & = \hat{f}(\lambda,t)e^{-a^2\lambda^2 t} \\ [\hat{u}(\lambda,t)e^{-a^2\lambda^2 t}]'_t & = \hat{f}(\lambda,t)e^{-a^2\lambda^2 t} \\ \hat{u}(\lambda,t)e^{-a^2\lambda^2 t} -\hat{u}(\lambda,0) & = \int _{0}^{t}\hat{f}(\lambda,u)e^{-a^2\lambda^2 u} \, du \\ \end{align}$$
$u(x,0) = 0 \implies \hat{u}(\lambda,0) = 0 $
which implies that :
$$\hat{u}(\lambda,t)= e^{a^2\lambda^2 t} \int _{0}^{t}\hat{f}(\lambda,u)e^{-a^2\lambda^2 u} \, du $$
the inverse :
$$u(x,t) = 2 \int_{0}^{+\infty} \hat{f}(\lambda,u)e^{a^2\lambda^2 [t-u]} \cos(\lambda x) \, du \, d \lambda$$
at this point I got stuck, how do I go on ?