[guided]The goal is to express entropy as the total amount of entropy dissipated by the Ornstein-Uhlenbeck flow. We start with a positive function $h$ bounded below by $a>0$ so that $\log(P_t h)$ is smooth and no singularity occurs in the quotient that will appear later.
Define
\begin{align*}
m := \int_{\mathbb{R}^n} h(x)\, d\gamma_n(x).
\end{align*}
The invariance of $\gamma_n$ under $P_t$ gives
\begin{align*}
\int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x)=m.
\end{align*}
Thus the second term in the entropy of $P_t h$ is independent of $t$:
\begin{align*}
\left(\int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x)\right)\log\left(\int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x)\right)=m\log m.
\end{align*}
Define the entropy curve $\Phi: [0,\infty) \to \mathbb{R}$ by
\begin{align*}
\Phi(t):=\operatorname{Ent}_{\gamma_n}(P_t h).
\end{align*}
Since $h$ and all its derivatives have polynomial growth, $P_t h$, $LP_t h$, and their derivatives have polynomial growth on compact $t$-intervals. The lower bound $h\geq a>0$ gives $P_t h\geq a$, so $\log(P_t h)$ has at most logarithmic polynomial growth. These bounds provide a Gaussian-integrable dominator for the derivative of the entropy integrand on compact $t$-intervals, so differentiating under the integral is justified by the [Dominated Convergence Theorem](/theorems/4).
We also need the evolution equation for the semigroup. For $t>0$, set $a_t:=e^{-t}$ and $b_t:=\sqrt{1-e^{-2t}}$. Differentiating the Gaussian formula for $P_t h$ under the integral sign gives
\begin{align*}
\partial_t P_t h(x)=\int_{\mathbb{R}^n}\nabla h(a_t x+b_t y)\cdot\left(-a_t x+\frac{a_t^2}{b_t}y\right)\,d\gamma_n(y).
\end{align*}
The term involving $y$ is converted into a second derivative by Gaussian integration by parts: for each coordinate $i$,
\begin{align*}
\int_{\mathbb{R}^n}y_i\partial_{x_i}h(a_t x+b_t y)\,d\gamma_n(y)=b_t\int_{\mathbb{R}^n}\partial_{x_i x_i}h(a_t x+b_t y)\,d\gamma_n(y).
\end{align*}
After summing over $i$, this gives
\begin{align*}
\partial_t P_t h(x)=a_t^2P_t(\Delta h)(x)-a_t x\cdot P_t(\nabla h)(x).
\end{align*}
On the other hand, differentiating $P_t h$ with respect to the spatial variable gives $\nabla P_t h=a_tP_t(\nabla h)$ and $\Delta P_t h=a_t^2P_t(\Delta h)$, so
\begin{align*}
\partial_tP_t h(x)=\Delta P_t h(x)-x\cdot\nabla P_t h(x)=L P_t h(x).
\end{align*}
Therefore the chain rule gives
\begin{align*}
\Phi'(t)=\int_{\mathbb{R}^n} (L P_t h)(x)\bigl(\log(P_t h(x))+1\bigr)\, d\gamma_n(x)-\frac{d}{dt}(m\log m).
\end{align*}
The last derivative is $0$. Also,
\begin{align*}
\int_{\mathbb{R}^n} (L P_t h)(x)\, d\gamma_n(x)=\frac{d}{dt}\int_{\mathbb{R}^n}P_t h(x)\, d\gamma_n(x)=0.
\end{align*}
Therefore the $+1$ term drops out, and we obtain
\begin{align*}
\Phi'(t)=\int_{\mathbb{R}^n} (L P_t h)(x)\log(P_t h(x))\, d\gamma_n(x).
\end{align*}
Now apply the Gaussian integration by parts identity for the generator $L$ to the functions $u=P_t h$ and $v=\log(P_t h)$. This identity is the ordinary coordinate integration by parts formula for the Gaussian density, with the term $-x\cdot\nabla u$ in $L$ accounting for the derivative of that density. The stated identity is first proved for polynomial-growth smooth test functions; here $v$ has logarithmic polynomial growth, so we justify the extension. Choose a smooth cutoff sequence $\rho_R: \mathbb{R}^n\to[0,1]$ with $\rho_R=1$ on $B(0,R)$, $\rho_R=0$ outside $B(0,2R)$, and $|\nabla\rho_R|\leq 2/R$, and set $v_R:=\rho_R\log(P_t h)$. Then $v_R\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$, so integration by parts applies to $u$ and $v_R$. The terms involving $\nabla\rho_R$ vanish as $R\to\infty$ because Gaussian tails dominate every polynomial-growth factor, and the remaining terms converge by dominated convergence. Thus the identity applies to $u=P_t h$ and $v=\log(P_t h)$:
\begin{align*}
\int_{\mathbb{R}^n} (Lu)(x)v(x)\, d\gamma_n(x)=-\int_{\mathbb{R}^n}\nabla u(x)\cdot\nabla v(x)\, d\gamma_n(x).
\end{align*}
Since
\begin{align*}
\nabla \log(P_t h)(x)=\frac{\nabla P_t h(x)}{P_t h(x)},
\end{align*}
we get
\begin{align*}
\Phi'(t)=-\int_{\mathbb{R}^n}\frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x).
\end{align*}
Finally, $P_t h$ converges to the constant $m$ as $t\to\infty$, and we spell out the domination needed for the entropy. Since $h$ has polynomial growth, choose constants $C>0$ and $k\in\mathbb{N}$ such that $|h(z)|\leq C(1+|z|^k)$ for all $z\in\mathbb{R}^n$. For every $t\geq0$ and $x\in\mathbb{R}^n$,
\begin{align*}
|P_t h(x)|\leq C\int_{\mathbb{R}^n}\left(1+|e^{-t}x+\sqrt{1-e^{-2t}}y|^k\right)\, d\gamma_n(y)\leq C_1(1+|x|^k),
\end{align*}
where $C_1>0$ depends only on $C,k,n$ and the Gaussian $k$-th moment. For fixed $x$, the expression $e^{-t}x+\sqrt{1-e^{-2t}}y$ converges to $y$. The integrand $h(e^{-t}x+\sqrt{1-e^{-2t}}y)$ is dominated, for all sufficiently large $t$, by $C_2(1+|x|^k+|y|^k)$ for a constant $C_2>0$, and this dominating function is integrable with respect to $\gamma_n(y)$. Therefore the [Dominated Convergence Theorem](/theorems/4) gives
\begin{align*}
P_t h(x)\to \int_{\mathbb{R}^n}h(y)\,d\gamma_n(y)=m.
\end{align*}
The lower bound $P_t h\geq a$ and the upper bound $|P_t h(x)|\leq C_1(1+|x|^k)$ imply that $(P_t h)\log(P_t h)$ is dominated by a fixed polynomial-growth function of $x$, hence by a $\gamma_n$-integrable function. The [Dominated Convergence Theorem](/theorems/4) with respect to $\gamma_n(x)$ gives
\begin{align*}
\lim_{t\to\infty}\Phi(t)=\operatorname{Ent}_{\gamma_n}(m)=0.
\end{align*}
For $S>0$, integrating the differential identity from $0$ to $S$ gives
\begin{align*}
\Phi(0)-\Phi(S)=\int_0^S \int_{\mathbb{R}^n}\frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x)\, d\mathcal{L}^1(t).
\end{align*}
The integrand is nonnegative, so the [Monotone Convergence Theorem](/theorems/509) applies as $S\to\infty$. Using $\Phi(S)\to0$, we obtain
\begin{align*}
\operatorname{Ent}_{\gamma_n}(h)=\int_0^\infty \int_{\mathbb{R}^n}\frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x)\, d\mathcal{L}^1(t).
\end{align*}[/guided]
