Evolution of Temperature in Time and Space in an Infinite Bar 

An infinitely long, uniform bar is perfectly insulated and one end
is kept at a constant temperature ($T_s$). How does its temperature
evolve in time and space?

Because the bar's cross-section is negligible W.R.T. its length, we can treat this as a quasi-$\text{1D}$ problem and Fourier's equation applies. We're looking for a function $T(x,t)$ that satisfies:
$$\partial_t T=\alpha \partial_{xx} T$$
and any boundary/initial conditions.
In addition we assume the bar's initial ($t=0$) temperature is $T_i$.
My question is inspired by this PSE question. It was initially framed for a constant heat flux $\mathbf{q}$ at $x=0$ but somewhat later the OP seems to have settled for the simpler case I presented above.
For a bar of finite length $L$ this becomes a fairly banal BVP (a second BC was used where the heat conduction at $x=L$ is $0$) with initial condition, and with solution of the form:
$$\boxed{T(x,t)=T_s+\displaystyle\sum_{n=1}^{\infty} B_n\exp\Big({-\alpha \Big(\frac{\pi n }{2L}\Big)^2 t}\Big)\cos\Big(\frac{\pi n x}{2L}\Big)}$$
For $n=1,2,3,...$
Determine the coefficients $B_n$ with the Fourier series:
$$B_n=\frac{2}{L}\int_0^LT_i\cos\Big(\frac{\pi n x}{2L}\Big)\mathrm{d}x$$
For a very long bar, i.e. $L\to \infty$, this tends to the question the OP asked.
At one point he then declared in the comments that he'd found the answer to his question on a page titled '2.2. Transient Conduction in Semi-Infinite Slab' (find the link provided by G.Smith in the comments). It's a very poor quality resource that posits the solution without any derivation whatsoever as being:
$$\boxed{\frac{T(x,t)-T_s}{T_i-T_s}=\mathrm{erf}\Big(\frac{x}{2\sqrt{\alpha t}}\Big)}$$
This looks like an extremely neat solution to a complex problem (at least to me) and I haven't the slightest idea how it was arrived at or whether it is even correct. The equation is dimensionally consistent, always a good sign of course.
On way forward could be to compare the solution for the the finite bar and the web page's solution for ever increasing values of $L$ and see if they converge. But that seems like an awful amount of work.
So my question really is: 'does anyone have a(n) idea(s) on how to tackle the $L=\infty$ problem?'
This page does confirm the validity of the $\mathrm{erf}$ solution.
 A: With thanks to 'Chet' in the comments it appears this problem (at least without convection) isn't really hard.
$$\partial_t T=\alpha \partial_{xx}T$$
Boundary conditions:
$$T(x,0)=T_i$$
$$T(0,t)=T_s$$
Introduce a Similarity Variable $\eta$:
$$\eta=\frac{x}{2\sqrt{\alpha t}}$$
This transforms the original PDE into an ODE:
$$\frac{\mathrm{d}^2T(\eta)}{\mathrm{d}\eta^2}=-2\eta \frac{\mathrm{d}T(\eta)}{\mathrm{d}\eta}\tag{1}$$
Transform also the boundary conditions:
$$T(x,0)=T_i\to T(\eta\to \infty)=T_i$$
$$T(0,t)=T_s\to T(\eta=0)=T_s$$
This then solves to:
$$\frac{T(x,t)-T_s}{T_i-T_s}=\mathrm{erf}(\eta)$$
$$\boxed{\frac{T(x,t)-T_s}{T_i-T_s}=\mathrm{erf}\Big(\frac{x}{2\sqrt{\alpha t}}\Big)}$$

Solving $(1)$ -
Substitute $z=T'(\eta)$ and $z'=T''(\eta)$, so that:
$$z'=-2\eta z$$
$$\frac{\mathrm{d}z}{z}=-2\eta$$
$$\ln z=-\eta^2+c$$
$$z=c_1\exp{(-\eta^2)}$$
$$T'(\eta)=c_1\exp{(-\eta^2)}$$
Integrate again:
$$T(\eta)=c_1\int \exp{(-\eta^2)}+c_2$$
Apply the BC and IV to determine $c_1$ and $c_2$.
