How to evaluate \[\begin{aligned} \int_{-\infty}^\infty e^{-s^2}s^4 ds & & ? \end{aligned}\]

Solution:

This clearly looks related to the Gaussian distribution. So we can start with \[\begin{aligned} \int_{-\infty}^\infty e^{-x^2} dx &= \sqrt{\pi} \end{aligned}\] Let \(x = \sqrt{\lambda} s\), so \(dx = \sqrt{\lambda}\, ds\) \[\begin{aligned} \int_{-\infty}^\infty e^{-x^2} dx &= \sqrt{\pi} \\ \int_{-\infty}^\infty e^{-\lambda s^2} \sqrt{\lambda}\, ds &= \sqrt{\pi} & \text{when }\lambda = 1 \\ \int_{-\infty}^\infty e^{-\lambda s^2}\, ds &= \frac{\sqrt{\pi}}{\sqrt{\lambda}} \\ \end{aligned}\] Apply differentiation with respect to \(\lambda\) twice, \[\begin{aligned} -\int_{-\infty}^\infty e^{-\lambda s^2} s^2\, ds &= -\frac{1}{2}\cdot\frac{\sqrt{\pi}}{\lambda^{3/2}} \\ \int_{-\infty}^\infty e^{-\lambda s^2} s^4\, ds &= \frac{3}{4}\cdot\frac{\sqrt{\pi}}{\lambda^{5/2}} \end{aligned}\] Set \(\lambda = 1\), \[\begin{aligned} \int_{-\infty}^\infty e^{-s^2} s^4\, ds &= \frac{3}{4}\sqrt{\pi} \end{aligned}\] \(\Box\)