Verifying the fundamental solution to the heat equation 
$$\displaystyle u(x,t) = \frac{1}{\sqrt{4 \pi t}}\int_{-\infty}^{\infty} e^{\frac{-(x-s)^2}{4t}}g(s)ds$$
Verify the above function satisfies $u_t - u_{xx} = 0$ for any function $g$.

I wrote $$\displaystyle u_t = -\frac{1}{4 \sqrt{\pi} t^{\frac{3}{2}}}\int_{-\infty}^{\infty} e^{\frac{-(x-s)^2}{4t}}g(s)ds + \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty} \frac{(x-s)^2}{4t^2} e^{\frac{-(x-s)^2}{4t}}g(s)ds$$ by the product rule.
and I got $$\displaystyle u_{xx} = \frac{1}{\sqrt{4 \pi t}}\int_{-\infty}^{\infty} \left (\frac{(x-s)^2}{4t^2}e^{\frac{-(x-s)^2}{4t}} - 2t e^{-\frac{(x-s)^2}{4t}} \right ) g(s)ds$$
However, I do not see how terms are supposed to cancel to get zero. Have I made a mistake?
 A: You just made an algebra error.  You should have
$$
\frac{\partial}{\partial x} \left[ e^{-\frac{(x-s)^2}{4 t}}\right] = -\frac{(x-s) e^{-\frac{(x-s)^2}{4 t}}}{2 t} 
$$
and so
$$
\frac{\partial^2}{\partial x^2} \left[ e^{-\frac{(x-s)^2}{4 t}}\right] = \frac{(x-s)^2 e^{-\frac{(x-s)^2}{4 t}}}{4 t^2} - \frac{e^{-\frac{(x-s)^2}{4 t}}}{2 t} \neq \frac{(x-s)^2 e^{-\frac{(x-s)^2}{4 t}}}{4 t^2} - 2t e^{-\frac{(x-s)^2}{4 t}}.
$$
If you correct the second term in your expression for $u_{xx}$ and pull a factor of $t$ out of the denominator of the integral, you will see that it is equal to $u_t$.
A: Let
$$
\varphi(s,t)=\frac{1}{\sqrt{4\pi t}}\exp\left(\frac{-s^{2}}{4t}\right).
$$
As you point out, the fundamental solution is
$$
u(x,t)=\int_{-\infty}^{\infty}g(s)\varphi(x-s,t)ds=[g*\varphi(\cdot,t)](x).
$$
Therefore, $u_t(\cdot, t) = g * \varphi_t(\cdot, t)$ and $u_{xx}(\cdot, t) = g * \varphi_{xx}(\cdot, t)$.
It follows that one need only show $\varphi_t = \varphi_{xx}$.
This can be established by direct computation (see here and here).
