= Solution
Complete the square in the <Fourier transform> exponent:
$$
-\lambda x^2-ikx=-\lambda\left(x+\frac{ik}{2\lambda}\right)^2-\frac{k^2}{4\lambda}.
$$
The supplied shifted <Gaussian integral> therefore gives
$$
\boxed{\widehat f(k)=\sqrt{\frac\pi\lambda}\exp\left(-\frac{k^2}{4\lambda}\right).}
$$
This is the <Fourier transform of a Gaussian>, with angular-wavenumber convention $e^{-ikx}$.
Take the spatial <Fourier transform> of the <heat equation>. Two <integrations by parts>, with decaying fields and first derivatives at infinity, give $\mathcal F[\psi_{xx}]=-k^2\widetilde\psi$. Thus each mode satisfies the <ordinary differential equation>
$$
\boxed{\partial_t\widetilde\psi(k,t)=-k^2\widetilde\psi(k,t).}
$$
Inserting the transformed <initial condition> and integrating the time equation gives
$$
\boxed{\widetilde\psi(k,t)=\sqrt{\frac\pi\lambda}\exp\left[-\left(t+\frac1{4\lambda}\right)k^2\right].}
$$
Finally set $A=t+1/(4\lambda)>0$. The <inverse Fourier transform>, evaluated by the same Gaussian identity, gives
$$
\begin{aligned}
\psi(x,t)&=\frac1{2\pi}\sqrt{\frac\pi\lambda}\int_{-\infty}^{\infty}e^{-Ak^2+ikx}dk\\
&=\frac1{2\sqrt{\lambda A}}e^{-x^2/(4A)}.
\end{aligned}
$$
Equivalently the <Gaussian initial data for the heat equation> evolve as
$$
\boxed{\psi(x,t)=\frac1{\sqrt{1+4\lambda t}}\exp\left[-\frac{\lambda x^2}{1+4\lambda t}\right],\qquad t>0.}
$$
The profile broadens and its peak decreases, while its total <integral> remains $\sqrt{\pi/\lambda}$, as required by <diffusion> without loss on the whole line. Direct differentiation verifies the equation and the initial limit.
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