Second moment of Silverman kernel

Question

I am learning kernel smoothing method, in the Wiki page , I find that second moment of Silverman kernel is equal to zero (i.e $\int u^{2} K(u) d u$), I am not sure how to derive this result, that integral look difficult to compute, it suffices to show that $$ \int_{0}^{\infty} x^{2} e^{-x / \sqrt{2}} \sin \left(\frac{x}{\sqrt{2}}+\frac{\pi}{4}\right) d x=0 $$ (Actually this equal to zero when I use WolframAlpha to verify), can any body help to solve this integral, original integral is
$$ \int_{-\infty}^{\infty} x^{2} \frac{1}{2} e^{-\frac{|x|}{\sqrt{2}}} \cdot \sin \left(\frac{|x|}{\sqrt{2}}+\frac{\pi}{4}\right)dx=0 $$

Answer 1

First let us substitute $y=x/\sqrt 2$. Then we have to show $$ J:= \int_0^\infty y^2\; e^{-y}\; (\sin y + \cos y)\; dy =0\ . $$ We use partial integration to reduce the degree of the polynomial $y^2$. $$ \begin{aligned} J &= \int_0^\infty y^2\; e^{-y}\; (\sin y + \cos y)\; dy \\ &= \int_0^\infty y^2\; (-e^{-y}\; \cos y)'\; dy \\ &= -\Big[ y^2\; e^{-y}\; \cos y \Big]_0^\infty + \int_0^\infty (y^2)'\; e^{-y}\; \cos y\; dy \\ &= \int_0^\infty 2y\; e^{-y}\; \cos y\; dy \\ &= \int_0^\infty y\; (e^{-y}\; (\sin y-\cos y))'\; dy \\ &= \Big[ y\; e^{-y}\; (\sin y-\cos y)\Big]_0^\infty - \int_0^\infty y'\; e^{-y}\; (\sin y-\cos y)\; dy \\ &= - \int_0^\infty e^{-y}\; (\sin y-\cos y)\; dy \\ &= \int_0^\infty (e^{-y}\; \sin y)'\; dy \\ &= \Big[ e^{-y}\; \sin y\Big]_0^\infty \\ &=0\ . \end{aligned} $$