Heat equation that depends on the viscousity constant $\epsilon$ I want to solve the following heat equation, for $\epsilon>0$:
$$\begin{cases}
      u_t-\epsilon u_{xx}=0 & x<0,\; t\in \mathbb{R^{+*}} \\
      u(0,x)= e^{-\frac{-x}{2\epsilon}}       & x<0 \\
      u(t,0)=1 & t\in \mathbb{R^{+*}} 
\end{cases}$$
And I wanna make sure is the explicit solution is:
$$u(t,x)=\frac{1}{\sqrt{4t\epsilon}}\int_{\mathbb{R^{-}}}u(0,s)e^{\frac{(x-s)^2}{4\epsilon}}ds $$
 A: Let $u(t,x)=T(t)X(x)$ ,
Then $T'(t)X(x)-\epsilon T(t)X''(x)=0$
$T'(t)X(x)=\epsilon T(t)X''(x)$
$\dfrac{T'(t)}{\epsilon T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{\epsilon T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-\epsilon ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(t,x)=\int_0^\infty C_1(s)e^{-\epsilon ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-\epsilon ts^2}\cos xs~ds$
$u(t,0)=1$ :
$\int_0^\infty C_2(s)e^{-\epsilon ts^2}~ds=T_0$
$C_2(s)=\delta(s)$
$\therefore u(t,x)=\int_0^\infty C_1(s)e^{-\epsilon ts^2}\sin xs~ds+\int_0^\infty \delta(s)e^{-\epsilon ts^2}\cos xs~ds$
$u(t,x)=\int_0^\infty C_1(s)e^{-\epsilon ts^2}\sin xs~ds+1$
$u(0,x)=e^\frac{x}{2\epsilon}$ :
$\int_0^\infty C_1(s)\sin xs~ds+1=e^\frac{x}{2\epsilon}$
$\mathcal{F}_{s,s\to x}\{C_1(s)\}=e^\frac{x}{2\epsilon}-1$
$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{e^\frac{x}{2\epsilon}-1\}=\dfrac{8\epsilon^2s}{\pi(4\epsilon^2s^2+1)}-\dfrac{2}{\pi s}$ (according to  http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf)
$\therefore u(t,x)=\dfrac{8\epsilon^2}{\pi}\int_0^\infty\dfrac{se^{-\epsilon ts^2}\sin xs}{4\epsilon^2s^2+1}~ds-\dfrac{2}{\pi}\int_0^\infty\dfrac{e^{-\epsilon ts^2}\sin xs}{s}~ds+1$
