= Solution
Put $s=\lambda t>0$. In the <heat-kernel convolution>, complete the real square as
$$
\alpha^2y^2+\frac{(x-y)^2}{4s}
=\left(\alpha^2+\frac1{4s}\right)
\left(y-\frac{x}{1+4s\alpha^2}\right)^2
+\frac{\alpha^2x^2}{1+4s\alpha^2}.
$$
The <Gaussian integral> therefore gives
$$
\boxed{\beta=\frac{\alpha}{\sqrt{1+4\lambda t\alpha^2}},\qquad
u(x,t)=\frac{\beta}{\sqrt\pi}e^{-\beta^2x^2}}.
$$
At $t=0$ the same formula gives $\beta=\alpha$. Equivalently, $\beta^{-2}=\alpha^{-2}+4\lambda t$: the <variance> of this normalized <Gaussian function> grows from $1/(2\alpha^2)$ to $1/(2\alpha^2)+2\lambda t$.
The initial profiles have unit integral and concentrate at zero as $\alpha\to\infty$. Precisely, for every smooth compactly supported <test function> $\varphi$,
$$
\int_{\mathbb R}\frac{\alpha}{\sqrt\pi}e^{-\alpha^2y^2}\varphi(y)\,dy
=\frac1{\sqrt\pi}\int_{\mathbb R}e^{-v^2}\varphi(v/\alpha)\,dv
\longrightarrow\varphi(0).
$$
Thus the limit is the <Dirac delta distribution>, rather than an ordinary function: pointwise the profiles tend to zero away from the origin and diverge at the origin. At every fixed $t>0$,
$$
\boxed{u(\cdot,0)\longrightarrow\delta_0\ \text{in distributions},\qquad
u(x,t)\longrightarrow\frac1{\sqrt{4\pi\lambda t}}e^{-x^2/(4\lambda t)}}.
$$
The limiting evolved profile is the <heat kernel> for a unit point source.
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