Showing that complicated Fourier integral satisfies heat equation I'm doing a chapter on Fourier analysis and I got a rather involved problem.
Given the function $u$,
\begin{align*}
u = \frac1{\sqrt{\pi t}} \int_{-\infty}^\infty \exp\left(-p^2-\frac{(x-p)^2}{t}\right)dp
\end{align*}
I want to show that it satisfies the (heat) equation $u_t = 4 u_{xx}$ and I want to compute its boundary values... Where do I start?
 A: In such a chapter, it was probably derived that
$$
               u(t,x)=\frac{1}{\sqrt{4\pi k t}}\int_{-\infty}^{\infty}f(y)e^{-(x-y)^2/4kt}dy
$$
is a solution of
$$
                       u_{t}(t,x) = ku_{xx}(t,x),\\
                   \lim_{t\downarrow 0}u(t,x) = f(x),
$$
where the limit in $t$ exists at any point $x$ where $f$ is continuous. The heat kernel $H(t,x)=\frac{1}{\sqrt{4\pi k t}}e^{-x^2/4kt}$ is a solution of $u_{t}(t,x)=ku_{xx}(t,x)$ that behaves as $\delta(x)$ as $t\downarrow 0$, which is why $\lim_{t\downarrow 0}u(t,x)=f(x)$ if $f$ is continuous.
You have a problem where $f(y)=e^{-y^2}$ and $k=1/4$. So your function $u$ satisfies
$$
                       u_{t}(t,x)=\frac{1}{4}u_{xx}(t,x),\\
                           u(0,x) = e^{-x^2}.
$$
Under fairly mild assumptions, the solution $u$ is unique. However, using the heat kernel, you can find another such solution.
If $u(t,x)$ is a solution of the heat equation, then $v(t,x)=u(t+a,x)$ is a solution of the same heat equation, but with a different initial function $v(0,x)=u(a,x)$. For example,
$$
                u(t,x) = \frac{1}{\sqrt{t}}e^{-x^2/t}
$$
is a solution of $u_{t}=\frac{1}{4}u_{xx}$, which means $v(t,x)=u(t+1,x)$ is a solution of $v_{t}=\frac{1}{4}v_{xx}$, and $v$ satisfies $v(0,x)=e^{-x^2}$. By uniqueness of solutions,
$$
    \frac{1}{\sqrt{t+1}}e^{-x^2/(t+1)}=v(t,x)=\frac{1}{\sqrt{\pi t}}\int_{-\infty}^{\infty}e^{-y^2}e^{-(x-y)^2/t}dy.
$$
