With the heat equation :
$$\frac{\partial T}{\partial t} = D ~\frac{\partial^2 T}{\partial z^2}$$
You got a kernel :
$$K(z,t) = \frac{1}{\sqrt{4\pi Dt}}~ e^{- \large \frac{z^2}{4Dt}}$$
such as :

$$T(z',t') = \int dz K(z'-z,t'-t) T(z,t)~~~~~~~~(0)$$

Noting $T(0,t') = T_0(t')$, we have :

$$T_0(t') = \int dz K(-z,t'-t) T(z,t)~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$$

This is true for all $t$, so this is true for $t=0$:

$$T_0(t') = \int dz K(-z,t') T(z,0)~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1')$$

It is simpler to consider $T_0(t')$ and $T(z,t)$ as the real part of complex quantities : $T_0(t') = \mathbb{Re}(\tilde T_0(t')), T(z,t)= \mathbb{Re}(\tilde T(z,t))$

Here, $\tilde T_0(t') = T_0 e^{ 2 i \pi \large \frac{t'-t_0}{t_d}}$, the relation between complex quantities is then :

$$\tilde T_0(t') = \int dz K(-z,t') \tilde T(z,0)~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$$

We are now using the identity :

$$\sqrt{\frac{2 \pi}{a}}e^{\large \frac{b^2}{2a}} = \int dz e^{- \large \frac{az^2}{2}+bz}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$$

With $a = \frac{1}{2Dt'}$ and $b = \sqrt{\large \frac{2 i \pi}{D t_d}}$, so that $\frac{b^2}{2a} = \frac{2i \pi t'}{t_d}$, this gives :

$$\sqrt{4 \pi Dt'}e^{\large \frac{2i \pi t'}{t_d}} = \int dz e^{- \large \frac{z^2}{4Dt'}+\sqrt{\large \frac{2 i \pi}{D t_d}}z}~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)$$

From this equation and equation $(2)$, we get :

$$\tilde T(z,0) = T_0 e^{\large \frac{-2i \pi t_0}{t_d}}e^{+ \sqrt{\large \frac{2 i \pi}{D t_d}}z} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~(5)$$

Now one can use the formula :

$$~\tilde T(z',t') = \int dz K(z'-z,t') \tilde T(z,0)~~~~~~~~~~~~~~~~~~~~~~~~(6)$$

This gives, using (4), (5), and some basic algebra :

$$~\tilde T(z',t') = T_0 e^{ 2 i \pi \large \frac{t'-t_0}{t_d}}e^{+ \sqrt{\large \frac{2 i \pi}{D t_d}}z'}~~~~~~~~~~~~~~~~~~~~~~~~(7)$$

The correct square root of $i$ is $-\frac{1+i}{\sqrt{2}}$ in order to have a decreasing exponential in $z$, taking the real part, we get the general expression for $T$, with $z' >0$:

$$~T(z',t') = T_0 ~ \cos( 2 \pi \frac{t'- t_0}{t_d} - \sqrt{ \frac{ \pi}{D t_d}}z')~e^{- \sqrt{\large \frac{\pi}{D t_d}}z'}~~~~~~~~~~~~~~~~~~~~~~~~(8)$$

At $t'=t_0$, we get :

$$~T(z',t_0) = T_0 ~\cos( \sqrt{ \frac{ \pi}{D t_d}}z')~e^{- \sqrt{\large \frac{\pi}{D t_d}}z'}~~~~~~~~~~~~~~~~~~~~~~~~(9)$$

The $z_{min}$ corresponding to the coolest temperature, is simply the first minimum of this function.

## Best Answer

It is easy. Observe that $$T[x,t] = \frac{f(x/s)}{s}$$ where $s = \sqrt{t}$ and $f(x) = T[x,1]$. Since $$\int_{\mathbb R} f(x) dx=1 $$ we also have $$\int_{\mathbb R} \frac{f(x/s)}{s} dx=1 $$ simply by means of a trivial change of variables, defining $z = x/s$ where $s>0$. Now take a bounded continuous function $g : \mathbb R \to \mathbb C$, with the aforementioned change of variables we have $$\int_{\mathbb R} \frac{f(x/s)}{s} g(x) dx= \int_{\mathbb R} \frac{f(x/s)}{s} g(sx/s) s dx/s = \int_{\mathbb R} f(z) g(sz) dz\:.$$ Therefore $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) dx = \lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} f(z) g(0) dz = g(0) \int_{\mathbb R} f(z) dz = g(0) 1 = g(0)\:.$$ In other words $$\lim_{t\to 0^+} \int_{\mathbb R}T[x,t]g(x) = g(0)\:. \tag{1}$$ The only crucial passage is $$\lim_{s\to 0^+} \int_{\mathbb R} f(z) g(sz) dz = \int_{\mathbb R} \lim_{s\to 0^+} f(z) g(sz) dz$$ A quite mild condition that guarantees the passage is that $g$ is bounded as already required (as consequence of

Lebesgue's dominated convergence theorem).I stress that (1) (where $g$ is a smooth compactly supported functions and we obtained the result with much weaker hypotheses) is one of the possible ways to rigorously state that $$T[x,0^+] = \delta(x)\:.$$ The heat kernel is used to construct the solution of the heat equation $g=g(x,t)$ out of the initial condition $g(x)$: $$g(x,t) = \int_{\mathbb R} T[x-y,t]g(y) dy $$ satisfies $$ \frac{∂g}{∂t} = D \frac{∂^2g}{∂x^2} $$ for $t>0$ with initial condition $$g(x,0) = g(x)\:.$$ The proof is immediate from (1).