I haven't had any experience with applying of FT to heat equation with source. But this popped up in an exercise.
Any help in the right direction would be great.
Consider:
$$\frac{\partial u(x,t)}{\partial t} = k\frac{\partial ^2u(x,t)}{\partial x^2} + Q(x,t) $$ subject to $$u(x,0)=f(x)$$ x lies on the infinite domain.
This reduces to a first order ODE in time $$ \frac{d}{dt}\hat{u}(s,t)=-ks^{2}\hat{u}(s,t)+\hat{Q}(s,t) \\ \frac{d}{dt}\left(e^{ks^{2}t}\hat{u}(s,t)\right)=e^{ks^{2}t}\hat{Q}(s,t) \\ e^{ks^{2}t}\hat{u}(s,t)-\hat{u}(s,0) = \int_{0}^{t}e^{ks^{2}t'}\hat{Q}(s,t')dt' \\ \hat{u}(s,t)=e^{-ks^{2}t}\hat{f}(s)+\int_{0}^{t}e^{-ks^{2}(t-t')}\hat{Q}(s,t')dt' $$ Now inverse transform: $$ u(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)e^{-ks^{2}t}e^{isx}ds+\int_{0}^{t} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{Q}(s,t')e^{-ks^{2}(t-t')}dsdt' $$ Then use the symmetry identity $$ \int_{-\infty}^{\infty} \hat{a}(s)b(s)ds = \int_{-\infty}^{\infty}a(x)\hat{b}(x)dx $$ along with the Fourier transform identity, $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-kts^{2}}e^{isx}ds=\frac{1}{\sqrt{4\pi kt}}e^{-tx^{2}/4kt}. $$ The final answer should be--assuming no typos--the following \begin{align} u(x,t) & =\frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^{\infty} e^{-(x-y)^{2}/4kt}f(y)dy \\ & +\int_{0}^{t}\frac{1}{\sqrt{4\pi k(t-t')}} \int_{-\infty}^{\infty}e^{-(x-y)^{2}/4k(t-t')}Q(y,t')dy dt' \end{align} In terms of the ordinary time propogater $T(t)$, $$ u(x,t) = T(t)f + \int_{0}^{t}T(t-t')Q(x,t')dt' $$