Case $1$: $t\geq0$
Let $\varphi(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=X''(x)T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore\varphi(x,t)=\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds$
Case $2$: $t\leq0$
Let $\varphi(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=X''(x)T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}=s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=s^2\\X''(x)-s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{ts^2}\\X(x)=\begin{cases}c_1(s)\sinh xs+c_2(s)\cosh xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore\varphi(x,t)=\int_0^\infty C_1(s)e^{ts^2}\sinh xs~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh xs~ds$
Hence $\varphi(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-ts^2}\sin xs~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos xs~ds&\text{when}~t\geq0\\\int_0^\infty C_1(s)e^{ts^2}\sinh xs~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh xs~ds&\text{when}~t\leq0\end{cases}$
Note that in this question we only care about $x\in\mathbb{R}$ . Since $\sin xs=-\sin(-xs)$ , $\cos xs=\cos(-xs)$ , $\sinh xs=-\sinh(-xs)$ and $\cosh xs=\cosh(-xs)$ , $\varphi(x,t)$ has alternative form of the solution:
$\varphi(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-ts^2}\sin|x|s~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos|x|s~ds&\text{when}~t\geq0\\\int_0^\infty C_1(s)e^{ts^2}\sinh|x|s~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh |x|s~ds&\text{when}~t\leq0\end{cases}$
Since $\lim\limits_{x\to\pm\infty}\sin|x|s$ , $\lim\limits_{x\to\pm\infty}\cos|x|s$ , $\lim\limits_{x\to\pm\infty}\sinh|x|s$ and $\lim\limits_{x\to\pm\infty}\cosh|x|s$ do not exist, we cannot determine directly whether $\lim\limits_{x\to\pm\infty}\varphi(x,t)=0$ holds or not.
However, when we do some change of variables:
$\varphi(x,t)=\begin{cases}\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)e^{-t\left(\frac{s}{|x|}\right)^2}\sin s~d\biggl(\dfrac{s}{|x|}\biggr)+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)e^{-t\left(\frac{s}{|x|}\right)^2}\cos s~d\biggl(\dfrac{s}{|x|}\biggr)&\text{when}~t\geq0\\\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)e^{t\left(\frac{s}{|x|}\right)^2}\sinh s~d\biggl(\dfrac{s}{|x|}\biggr)+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)e^{t\left(\frac{s}{|x|}\right)^2}\cosh s~d\biggl(\dfrac{s}{|x|}\biggr)&\text{when}~t\leq0\end{cases}$
$\varphi(x,t)=\begin{cases}\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{-\frac{ts^2}{x^2}}\sin s}{|x|}ds+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{-\frac{ts^2}{x^2}}\cos s}{|x|}ds&\text{when}~t\geq0\\\int_0^\infty C_1\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{\frac{ts^2}{x^2}}\sinh s}{|x|}ds+\int_0^\infty C_2\biggl(\dfrac{s}{|x|}\biggr)\dfrac{e^{\frac{ts^2}{x^2}}\cosh s}{|x|}ds&\text{when}~t\leq0\end{cases}$
Since $\lim\limits_{x\to\pm\infty}\dfrac{e^{-\frac{ts^2}{x^2}}}{|x|}=0$ and $\lim\limits_{x\to\pm\infty}\dfrac{e^{\frac{ts^2}{x^2}}}{|x|}=0$ ,
$\therefore\varphi(x,t)=\begin{cases}\int_0^\infty C_1(s)e^{-ts^2}\sin|x|s~ds+\int_0^\infty C_2(s)e^{-ts^2}\cos|x|s~ds&\text{when}~t\geq0\\\int_0^\infty C_1(s)e^{ts^2}\sinh|x|s~ds+\int_0^\infty C_2(s)e^{ts^2}\cosh |x|s~ds&\text{when}~t\leq0\end{cases}$ automatically satisflies $\lim\limits_{x\to\pm\infty}\varphi(x,t)=0$ .
The remaining problem is how to substitute $\varphi(x,0)=f_0(x)$ to eliminate nicely on some of the $C_1(s)$ and $C_2(s)$ .