I have the diffusion equation
$$\frac{\partial u}{\partial t} = \mu\frac{\partial^2u}{\partial y^2} $$
in the plane $y\ge 0$ with $u(0,t)=U$ and $u(\infty,t) = 0$ and am trying to solve it via a self similar solution using $\eta = \frac{y}{\sqrt{4\mu t}}$. The question asks to show that at time $t = T$ a fluid particle a distance $y = h$ from the plane has speed
$$\frac{2U}{\sqrt{\pi}}\int_{\alpha}^\infty e^{-s^2} ds, \alpha^2 = \frac{h^2}{4\mu T} $$
My attempt at a solution is as follows:
$$\eta = \frac{y}{\sqrt{4\mu t}},u = f(\eta), \eta_t = \frac{-y}{2(4\mu)^{\frac{1}{2}}t^{\frac{3}{2}}}=\frac{-\eta}{2t},\eta_y=\frac{1}{(4\mu t)^{\frac{1}{2}}} $$
Substituting in and cancelling:
$$-\eta f'=\frac{1}{2}f'' \\ \implies\int\frac{f''}{f'} \mathrm{d\eta} =\int -2\eta\mathrm{d\eta} \\ \implies f' = ke^{-\eta^2} \\ \implies u(y,t) = f(\eta) = k\int e^{-\eta^2} + c$$
And from here I get lost, as I do not understand why the limits have been chosen the way they have in the question (I.e. from $\alpha$ to $\infty$ rather than from $0$ to $\alpha$), and I also don't understand how there is a $\frac{2}{\sqrt{\pi}}$ in the answer as well as a Gaussian integral, as if it has been evaluted (yet it is still there). Also I am unsure if I can apply the boundary conditions inside an integral. I can see the final answer has the form of a complementary error function but I don't understand how to get there. Many Thanks.