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 \(\lambda = 1\), \[\begin{aligned} \int_{-\infty}^\infty e^{-s^2} s^4\, ds &= \frac{3}{4}\sqrt{\pi} \end{aligned}\] \(\Box\)

A fun practice of double integration by parts. Evaluate: \[\begin{aligned} -\int_R \frac{\partial^2}{\partial y^2}\left(\frac{t}{t^2 + y^2}\right)\cos y\,dy \end{aligned}\]

Solution:

\[\begin{aligned} -\int_R \frac{\partial^2}{\partial y^2}\left(\frac{t}{t^2 + y^2}\right)\cos y\,dy &= -\int_R \frac{\partial}{\partial y}\left(\frac{\partial}{\partial y}\left(\frac{t}{t^2 + y^2}\right)\right)\cos y\,dy \\ &= -\frac{\partial}{\partial y}\left(\frac{t}{t^2 + y^2}\right)\bigg\vert_{y=-\infty}^{\infty} - \int_R \frac{\partial}{\partial y}\left(\frac{t}{t^2 + y^2}\right) \sin y\,dy \\ &= -\int_R \frac{\partial}{\partial y}\left(\frac{t}{t^2 + y^2}\right) \sin y\,dy \\ &= -\left(\frac{t}{t^2 + y^2}\right) \sin y \bigg\vert_{y=-\infty}^{\infty} + \int_R \left(\frac{t}{t^2 + y^2}\right) \cos y\,dy \\ &= \int_R \frac{t\cos y}{t^2 + y^2} \,dy \end{aligned}\] \(\Box\)

Ever wonder how the Euler's Factorial Integral \[\begin{aligned} \int_0^\infty x^n e^{-x} dx = n! \end{aligned}\] can be obtained from \(\displaystyle \int_0^\infty e^{-x} dx \) ?

A great practice of repeated integration by parts: \[\begin{aligned} \int_0^\infty e^{-x} dx = -e^{-x}\big\vert_0^\infty &= 1 \\ \int_0^\infty e^{-x} dx = \int_0^\infty e^{-x} x' dx = x e^{-x}\big\vert_0^\infty + \int_0^\infty x e^{-x} dx &= \int_0^\infty x e^{-x} dx \\ \int_0^\infty (\frac{x^2}{2})' e^{-x} dx = \frac{x^2}{2} e^{-x}\big\vert_0^\infty + \frac{1}{2} \int_0^\infty x^2 e^{-x} dx &= \frac{1}{2} \int_0^\infty x^2 e^{-x} dx \\ \frac{1}{2} \int_0^\infty (\frac{x^3}{3})' e^{-x} dx = \frac{1}{2}\cdot \frac{x^3}{3} e^{-x}\big\vert_0^\infty + \frac{1}{2\cdot 3} \int_0^\infty x^3 e^{-x} dx &= \frac{1}{3!} \int_0^\infty x^3 e^{-x} dx \\ &\vdots \\ \frac{1}{(n-1)!} \int_0^\infty (\frac{x^n}{n})' e^{-x} dx = \frac{1}{(n-1)!} \frac{x^n}{n} e^{-x}\big\vert_0^\infty + \frac{1}{(n-1)!} \frac{1}{n} \int_0^\infty x^n e^{-x} dx &= \frac{1}{n!} \int_0^\infty x^n e^{-x} dx \\ \therefore \frac{1}{n!} \int_0^\infty x^n e^{-x} dx &= 1 \\ \int_0^\infty x^n e^{-x} dx &= n! \end{aligned}\] \(\Box\)

But wait. We can do better than this!

See Differentiating under the integral.