I have difficulty with the following calculation : $\lVert f'\rVert_{L^2}^{2}$ with $f(x)=\frac{1}{(2\pi\sigma^{2})^{1/2}}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}$
We have $f'(x)=\frac{-(x-\mu)}{(2\pi)^{1/2}\sigma^{3}}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}$
Then $(f'(x))^{2}=\frac{(x-\mu)^{2}}{2\pi\sigma^{6}}e^{\frac{-(x-\mu)^{2}}{\sigma^{2}}}$
$\lVert f'\rVert_{L^2}^{2}=\int_{-\infty}^{+\infty}(f'(x))^{2}dx=\int_{-\infty}^{+\infty}\frac{(x-\mu)^{2}}{2\pi\sigma^{6}}e^{\frac{-(x-\mu)^{2}}{\sigma^{2}}}dx=\frac{1}{2\pi\sigma^{6}}\int_{-\infty}^{+\infty}(x-\mu)^{2}e^{\frac{-(x-\mu)^{2}}{\sigma^{2}}}dx=\frac{1}{2\pi\sigma^{6}}\frac{(\sigma^{2}\pi)^{1/2}}{(\sigma^{2}\pi)^{1/2}}\int_{-\infty}^{+\infty}(x-\mu)^{2}e^{\frac{-(x-\mu)^{2}}{\sigma^{2}}}dx=\frac{1}{2(\pi)^{1/2}\sigma^{5}}\int_{-\infty}^{+\infty}(x-\mu)^{2}\frac{e^{\frac{-(x-\mu)^{2}}{\sigma^{2}}}}{(\sigma^{2}\pi)^{1/2}}dx$
where the last integral corresponds to the variance of $\mathcal{N}\left(\mu,\frac{\sigma^2}{2}\right)$ thus we have
$\lVert f'\rVert_{L^2}^{2}=\frac{1}{4(\pi)^{1/2}\sigma^{3}}$
Is this seems correct ?
Thank you a lot !
Thank you to Lukas for his help !