Show that the solution to the problem $$\frac{\partial^2 u}{\partial x^2}=a^{-2}\frac{\partial u}{\partial t}$$ subject to $u_x(0,t)=-f(t)$ and $u(x,t)\to 0$ as $x\to \infty$ and $u(x,0)=0$ where $0<x<\infty$ and $t>0$ is given by $$u(x,t)=\frac{a}{\sqrt{\pi}}\int_0^t \frac{f(\tau)}{\sqrt{t-\tau}}\exp\bigg[-\frac{x^2}{4a^2(t-\tau)}\bigg]d\tau$$
Solution is fairly easy. I took Laplace transform on both sides of given equation w.r.t. $t$ to get $$\frac{d^2}{dx^2}U(x,s)=a^{-2}(sU(x,s)-u(x,0)) \\ \implies U(x,s)=Ae^{\frac{x\sqrt{s}}{a}}+Be^{-\frac{x\sqrt{s}}{a}}$$ where $U(x,s)=\mathcal{L}(u(x,t);s)$. Now using $u(x,t)\to 0$ as $x\to \infty$ we have $A=0$. So $U(x,s)=Be^{-\frac{x\sqrt{s}}{a}}$. Finally taking Laplace of $u_x(0,t)=-f(t)$ we get $$U_x(0,s)=-F(s)$$ So we have $$B=\frac{a}{\sqrt{s}}F(s)$$ where $F(s)=\mathcal{L}(f(t);s)$. Finally we have $$U(x,s)=\frac{a}{\sqrt{s}}F(s)e^{-\frac{x\sqrt{s}}{a}}$$ Clearly an application of convolution would conclude the answer, but problem is we have product of $3$ functions here, I tried at first that $$u(x,t)=\mathcal{L}^{-1}\bigg(\frac{a}{\sqrt{s}}F(s)e^{-\frac{x\sqrt{s}}{a}};t\bigg)=f(t)*\mathcal{L}^{-1}\bigg(\frac{a}{\sqrt{s}}e^{-\frac{x\sqrt{s}}{a}};t\bigg)$$ where "$*$" denotes the convolution. But I don't have the formula of $$\mathcal{L}^{-1}\bigg(\frac{a}{\sqrt{s}}e^{-\frac{x\sqrt{s}}{a}};t\bigg)$$ Is there an easy formula of this which will conclude the answer? Thanks in advance.