A semi-infinite string with one end fixed at the origin, is strectched along the positive half of the $x$ axis and released at rest from a position $y=f(x)$ $(x\geq 0)$. Derieve the expression
$$y(x,t)=\frac{2}{\pi}\int_0^{\infty} \cos\alpha(at) \sin\alpha(x)\int_{0}^{\infty} f(s)\sin\alpha(s) \,\mathrm ds\,\mathrm da$$
for the tranverse displacement. Let $F(x)(-\infty<x<\infty)$ denote the odd extention of $f(x)$ and show how this result reduces to
$$y(x,t)=\frac{1}{2}[F(x+at)+F(x-at)$$
What i tried
The expression
$$y(x,t)=\frac{1}{2}[F(x+at)+F(x-at)$$
which i know immediately as the De-Alembert solution of the wave equation
My boundary conditions are
$$u_{t}(x,t)=ku_{xx}(x,t)$$
$$u(t,0)=0$$
$$u(x,0)=f(x)$$ Is my boundary conditions correct?
THen using seperation of variables, i got
$$X''(x)+\lambda X(x)=0$$
$$X(0)=0$$
and
$$T'(t)+\lambda kT(t)=0$$
The eigenvalues are
$$\lambda=\alpha^2$$
Eigenvectors are $$T(t)=\exp(-\alpha^2 kt)$$
$$u(x,t)=\int_{0}^\infty \beta(\alpha)\exp(-\alpha^2 kt)\sin(\alpha (ax) da$$ Im not sure of how to continue from here. Could anyone explain every step of the problem for me.Thanks