更改

\int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{f([-\frac{L}{2},\frac{L}{2}])}p(x)p(y|x)\ln p(y|x)dydx&\approx \int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{-\infty}^{\infty}p(x)p(y|x)\ln p(y|x)dydx\\

\int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{f([-\frac{L}{2},\frac{L}{2}])}p(x)p(y|x)\ln p(y|x)dydx&\approx \int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{-\infty}^{\infty}p(x)p(y|x)\ln p(y|x)dydx\\

&=\int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{-\infty}^{\infty}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\ln\left[\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\right]dydx\\

&=\int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{-\infty}^{\infty}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\ln\left[\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\right]dydx\\

&=\ln(\frac{L}{\sigma\cdot\sqrt{2\pi e}})

&=\ln(\frac{1}{\sigma\cdot\sqrt{2\pi e}})

\end{aligned}

\end{aligned}

</math>

</math>