Let us consider the heat equation on the quarter plane \begin{align*} \frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2},\quad t>0,\;x>0 \end{align*} with boundary condition $u(0,t) = A$ and initial condition $u(x,0) = 0$.
Now we introduce $w(x,t) = u(x,t)-A$ where $A\in\mathbb{R}$, $u(x,t)$ is a solution to the heat equation and I've shown that $w(x,t)$ is also a solution to the heat equation. Also we have boundary and initial conditions \begin{align*} w(0,t)&=u(0,t)-A=A-A=0\\ w(x,0)&=u(x,0)-A=0-A=-A. \end{align*}
By some formula from the book we have that \begin{align*} w(x,t)&=\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}w(s,0)\left(e^{-(x-s)^2/4t}-e^{-(x+s)^2/4t}\right)ds\\ &=\frac{-A}{2\sqrt{\pi t}}\int_0^{\infty}\left(e^{-(x-s)^2/4t}-e^{-(x+s)^2/4t}\right)ds. \end{align*}
Furthermore, I've shown that \begin{align*} u(x,t)=A\left(1+\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}\left(e^{-(x+s)^2/4t}-e^{-(x-s)^2/4t}\right)ds\right). \end{align*} Now I want to prove that $u(x,t) = 2AN\left(\frac{-x}{\sqrt{2t}}\right)$.
Let's begin with noting that \begin{align}\label{inte1} \frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-s^2}ds=\frac{1}{4}\frac{2}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-s^2}ds\overset{\text{def}}{=}\frac{\text{erf}(x)}{4}+C\quad(1) \end{align} and since $y=e^{-s^2}$ is symmetric in the $y$-axis, we also say \begin{align}\label{inte2} \frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}e^{-s^2}ds=\frac{1}{2\sqrt{\pi}}\int_{-\infty}^{0}e^{-s^2}ds=\frac{\text{erf}(x)}{8}+C. \end{align} Let us choose $C=0$ for a moment. So far, so good, I hope. Moreover, we note that in the integral equation (1) we can replace $s$ by $(x-s)$ or $(x+s)$ without the integral changing its value because whether $s$ ranges from $-\infty$ to $\infty$, for exmaple, or $(x-s)$ ranges from $-\infty$ to $\infty$ does not give another value to the integral; i.e. \begin{align}\label{inte3} \frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-s^2}ds=\frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-(x-s)^2}ds=\frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-(x+s)^2}ds. \end{align} I do not know whether this equation above is right. Now I try to go to the desired result and end up with question marks: \begin{align*} u(x,t)&=A\left(1+\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}\left(e^{-(x+s)^2/4t}-e^{-(x-s)^2/4t}\right)ds\right)\\ &=A\left(\frac{1}{2}+\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}e^{-(x+s)^2/4t}ds+\frac{1}{2}-\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}e^{-(x-s)^2/4t}ds\right)\\ &=A\left(\frac{1}{2}+\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}e^{-(x+s)^2/4t}ds+\frac{1}{2}+\frac{1}{2\sqrt{\pi t}}\int_0^{\infty}e^{-(x+s)^2/4t}ds\right)\quad(??)\\ &= \end{align*}
I have no idea how to turn that $1$ into some integral. Maybe it is true that \begin{align*} \frac{1}{2\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-s^2}ds=\frac{1}{2}? \end{align*} Does anyone have some suggestion?
Let $u(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=X''(x)T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds$
$u(0,t)=A$ :
$\int_0^\infty C_2(s)e^{-ts^2}~ds=A$
$C_2(s)=A\delta(s)$
$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty A\delta(s)e^{-ts^2}\cos xs~ds$
$u(x,t)=\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+A$
$u(x,0)=0$ :
$\int_0^\infty C_1(s)\sin xs~ds+A=0$
$\mathcal{F}_{s,s\to x}\{C_1(s)\}=-A$
$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{-A\}=-\dfrac{2A}{\pi s}$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf)
$\therefore u(x,t)=A-\dfrac{2A}{\pi}\int_0^\infty\dfrac{e^{-ts^2}\sin xs}{s}~ds=A~\text{erfc}\left(\dfrac{x}{2\sqrt t}\right)$