$\displaystyle I=\int_{-\infty}^{+\infty} [f''(x)]^2 \, dx=\int_{-\infty}^{+\infty} \frac{(x^2-\sigma^2)^2}{\sigma^8} f^2 \ dx=\frac{1}{\pi \sigma^{10}} \int_0^{+\infty} (x^4-2 \sigma^2 x^2 +\sigma^4) \ e^{-\frac{x^2}{\sigma^2}} \ dx $

$\displaystyle \text{Let } u=\frac{x^2}{\sigma^2} \text{, then } I= \frac{1}{2 \pi \sigma^5} \int_0^{+\infty} (u^{\frac{3}{2}}-2u^{\frac{1}{2}} +u^{-\frac{1}{2}}) \ e^{-u} \ du $$\displaystyle = \frac{1}{2 \pi \sigma^5} \left[\Gamma(\tfrac{5}{2})-2\Gamma(\tfrac{3}{2})+\Gamma(\tfrac{1}{2})\right]=\frac{3}{8}\pi^{-1/2}\sigma ^{-5}$