= Solution
Take $\lambda>0$ and $t>0$, the <positive diffusivity requirement for the forward heat kernel>. The PDF does not explicitly specify the sign of $\lambda$; the stated real <Gaussian function> formula requires this assumption. For a direct calculation, initially assume sufficiently smooth and rapidly decreasing data, so that the <Fourier transform>, its inverse and the following integrations are justified. The resulting <heat-kernel convolution> extends to wider classes of initial data.
Using the stated <Fourier transform> convention and <integration by parts>,
$$
\widetilde{u_{xx}}(k,t)=-k^2\widetilde u(k,t),\qquad
\partial_t\widetilde u=-\lambda k^2\widetilde u.
$$
The transformed <ordinary differential equation> has solution $\widetilde u(k,t)=e^{-\lambda k^2t}\widetilde u_0(k)$. <Fourier inversion> then gives
$$
u(x,t)=\int_{-\infty}^{\infty}u_0(y)K_t(x-y)\,dy,\qquad
K_t(s)=\frac1{2\pi}\int_{-\infty}^{\infty}e^{-\lambda tk^2+iks}\,dk.
$$
Complete the square:
$$
-\lambda tk^2+iks
=-\lambda t\left(k-\frac{is}{2\lambda t}\right)^2-\frac{s^2}{4\lambda t}.
$$
Part (a), after scaling by $\sqrt{\lambda t}$, evaluates the remaining <Gaussian integral> even though its shift is imaginary. Therefore
$$
K_t(s)=\frac{e^{-s^2/(4\lambda t)}}{\sqrt{4\pi\lambda t}},\qquad
\boxed{u(x,t)=\frac1{\sqrt{4\pi\lambda t}}\int_{-\infty}^{\infty}
e^{-(x-y)^2/(4\lambda t)}u(y,0)\,dy}.
$$
The <heat kernel> has integral one and is an <approximate identity> as $t\downarrow0$, which recovers the initial condition. Mere decay at spatial infinity does not by itself imply every Fourier integral is absolutely convergent; the regularity assumption above supplies a rigorous setting for the derivation.
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