Problem
For $N \geq 1$, we consider the Heat Kernel:
$$
\Phi (x,t) =
\begin{cases}
\dfrac{1}{(4\pi t)^{N/2}} e^{-|x|^{2}/4t} &\text{ for }x \in \mathbb{R}^{N}, t > 0\\
\\
0&\text{ for }x \in \mathbb{R}^{N}, t < 0
\end{cases}.
$$
We define the set $E(1) = \{ (y,s) \in \mathbb{R}^{N+1} : s \leq 0, \Phi (-y,-s) \geq 1 \}$: our goal is to show the following integral equality:
$$
\iint\limits_{E(1)} \frac{|y|^2}{s^2} \mathrm{d}y\mathrm{d}s = 4.
$$
Attempted Answer
We set $N = 1$ to start, and investigate when $(y,s) \in E(1)$. We know $(y,s) \in E(1) \implies s \leq 0 $, then
$$
\begin{split}
\Phi (-y,-s) &= \dfrac{1}{(4\pi [-s])^{1/2}} e^{-|y|^{2}/4[-s]} \geq 1\\
&\iff e^{-|y|^{2}/4[-s]} \geq (4\pi [-s])^{1/2}\\
\\
&\iff e^{-|y|^{2}/2[-s]} \geq (4\pi [-s])\\
\\
&\iff e^{-|y|^{2}} \geq (4\pi [-s])^{2[-s]}\\
\\
&\iff
\begin{split}
-|y|^{2} & \geq \ln[(4\pi [-s])^{2[-s]}] \\
& = 2[-s]\ln(4\pi [-s])
\end{split}\\
\\
&\iff |y|^{2} \leq (2s)\ln(-4\pi s)\\
\\
&\iff |y| \leq [(2s)\ln(-4\pi s)]^{1/2},
\end{split}
$$
and finally
$$
(2s)\ln(-4\pi s) \geq 0,
$$
so we have a bound for $y$ in terms of $s$.
We next note that, since $s \leq 0$, we have $(2s)\ln(-4\pi s) \geq 0 \iff s \in \big[\frac{-1}{4\pi}, 0\big)$, thus we can write the integral as:
$$
\iint\limits_{E(1)} \frac{|y|^2}{s^2} \mathrm{d}y\mathrm{d}s = \int\limits^{0}_{\frac{-1}{4\pi}} \int\limits^{[(2s)\ln(-4\pi s)]^{1/2}}_{-[(2s)\ln(-4\pi s)]^{1/2}} \frac{|y|^2}{s^2} \mathrm{d}y\mathrm{d}s.
$$
However, the final form of this integral I calculated is the following:
$$
\int^{0}_{\frac{-1}{4\pi}} \frac{2^{5/2}}{3} s^{-1/2} \ln(-4 \pi s)^{3/2} \mathrm{d}s
$$
This, according to an integral calculator, does not work out to anything like the answer we need. Could someone please point out where I might have gone wrong?
Any advice as to how I could then generalise to the dimension $N$ case would also be appreciated!
In the paper Zujin Zhang Mean-value property of the heat equation the related calculations are carried out in detail.