Consider the Cauchy problem $$\begin{cases} u_{t}-u_{xx}=0 \ \ &\text{for} \ \ -\infty\lt x\lt +\infty , t\gt0\\ u(x,0)=\sin x \ \ \\ \end{cases}$$
I was doing like this : $$u=\frac{1}{2 \sqrt{\pi t}} \int_{-\infty}^{\infty}e^{-\frac{(x-y)^2}{4t}}\sin y \text dy $$ After the substitution $m=\frac{x-y}{2\sqrt{t}}$ , I get $$u=\frac{1}{ \sqrt{\pi }} \int_{-\infty}^{\infty}e^{-m^2}\sin (2\sqrt t m+x) \text dm $$ Now I can't move forward. How to simplify this or is there any problem in applying the formula.
Instead of applying the fundamental solution (which should work using the hint given above), you can also use that the initial conditions are periodic and solve the equation using a Fourier series ansatz $\sum_{n=1}^\infty a_n(t) \sin(nx)$. You then get an ODE for $a_n(t)$ that is easily solved (and because of your initial conditions, all but $a_1(t)$ are identically zero. The answer you get from this is $u(t)=e^{-t}\sin x$, which is the unique bounded solution.