小テク集2:Kernel swapping

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}\mathrm{d}x& ={\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(1)}{x^1}\sin x\mathrm{d}x\\ & ={\int }_0^{\mathrm{\infty }}\mathcal{L}[1](x)\sin x\mathrm{d}x\\ & ={\int }_0^{\mathrm{\infty }}1\mathcal{L}[\sin x](y)\mathrm{d}y\\ & ={\int }_0^{\mathrm{\infty }}\frac{1}{1+y^2}\mathrm{d}y\\ & =\frac{\pi }{2}.\end{array}$$

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\sin x^2\mathrm{d}x& ={\int }_0^{\mathrm{\infty }}\frac{1}{2}{u}^{-1/2}\sin u\mathrm{d}u(x^2=u)\\ & =\frac{1}{2\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(1/2)}{u^{1/2}}\sin u\mathrm{d}u\\ & =\frac{1}{2\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\mathcal{L}[t^{-1/2}](u)\sin u\mathrm{d}u\\ & =\frac{1}{2\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{t}^{-1/2}\mathcal{L}[\sin u](t)\mathrm{d}t\\ & =\frac{1}{2\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{t^{-1/2}}{1+t^2}\mathrm{d}t\\ & =\frac{1}{4\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{s^{1/4-1}}{1+s}\mathrm{d}s(t^2=s)\\ & =\frac{1}{4\sqrt{\pi }}\frac{\pi }{\sin \left(\pi /4\right)}\\ & =\frac{\sqrt{2\pi }}{4}.\end{array}$$

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}\cos x\mathrm{d}x& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{\pi }}\mathcal{F}[e^{-{\xi }^2}](2x)\cos x\mathrm{d}x(\mathrm{注}\mathrm{：}\mathcal{F}[e^{-t^2}](\omega )=\sqrt{\pi }{e}^{-{\omega }^2/4})\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{\pi }}{e}^{-{\xi }^2}\mathcal{F}[\cos x](2\xi )\mathrm{d}\xi \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{\pi }}{e}^{-{\xi }^2}\pi [\delta (2\xi -1)+\delta (2\xi +1)]\mathrm{d}\xi \\ & =\frac{\sqrt{\pi }}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{\xi }^2/4}[\delta (\xi -1)+\delta (\xi +1)]\mathrm{d}\xi (\delta (ax-b)=\frac{1}{|a|}\delta (x-\frac{b}{a})\mathrm{で}\mathrm{も}\mathrm{よ}\mathrm{い}.)\\ & =\sqrt{\pi }{e}^{-1/4}.\end{array}$$