이것이 바로 계산 잉여력

쉬운 것도 어렵게 하라! \begin{align*} & \int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}\;dx \\ &=\int_{-\infty}^{\infty}\cos x\left(\int_{0}^{\infty}e^{-(1+x^2)t}\;dt\right)dx\\ &=\int_{0}^{\infty}e^{-t}\left(\int_{-\infty}^{\infty}\cos x\,e^{-tx^2}\;dx\right)dt\\ &=\int_{0}^{\infty}e^{-t}\left(\int_{-\infty}^{\infty}e^{-tx^2+ix}\;dx\right)dt\\ &=\int_{0}^{\infty}e^{-t-\frac{1}{4t}}\left(\int_{-\infty}^{\infty}e^{-t\left(x-\frac{i}{2t}\right)^2}\;dx\right)dt\\ &=\int_{0}^{\infty}e^{-t-\frac{1}{4t}}\left(\int_{-\infty}^{\infty}\left( e^{-tx^2} + \int_{0}^{\frac{1}{2t}} i2t(x-is) e^{-t(x-is)^2} \; ds \right) \;dx\right)dt\\ &=\int_{0}^{\infty}e^{-t-\frac{1}{4t}}\left(\int_{-\infty}^{\infty} e^{-tx^2} \;dx -i \int_{0}^{\frac{1}{2t}} \left[ e^{-t(x-is)^2} \right]_{-\infty}^{\infty} \;ds \right)dt\\ &=\int_{0}^{\infty}e^{-t-\frac{1}{4t}}\left(\int_{-\infty}^{\infty} e^{-tx^2} \;dx \right)dt\\ &=\left(\int_{-\infty}^{\infty} e^{-x^2} \;dx \right)\left( \int_{0}^{\infty}\frac{1}{\sqrt{t}}e^{-t-\frac{1}{4t}}dt \right) \qquad (x\sqrt{t} \mapsto x)\\ &=\sqrt{2}\left(\int_{-\infty}^{\infty} e^{-x^2} \;dx \right)\int_{0}^{\infty}e^{-\frac{1}{2}\left(u^2+\frac{1}{u^2}\right )}\;du\qquad(t=u^2/2)\\ &=\frac{1}{\sqrt{2}}\left(\int_{-\infty}^{\infty} e^{-x^2} \;dx \right)\left( \int_{0}^{\infty}e^{-\frac{1}{2}\left(u^2+\frac{1}{u^2}\right )}\;du + \int_{0}^{\infty}\frac{1}{u^2}e^{-\frac{1}{2}\left(u^2+\frac{1}{u^2}\right )}\;du \right)\qquad(u\mapsto1/u)\\ &=\frac{1}{\sqrt{2}}\left(\int_{-\infty}^{\infty} e^{-x^2} \;dx \right)\left(\int_{0}^{\infty}\left(1 + \frac{1}{u^2} \right) e^{-\frac{1}{2}\left(u-\frac{1}{u}\right)^2-1}\;du\right)\\ &=\frac{1}{e}\left(\int_{-\infty}^{\infty} e^{-x^2} \;dx \right)\left(\int_{-\infty}^{\infty}e^{-y^2}\;dy\right)\qquad(\sqrt{2}y=u-u^{-1})\\ &=\frac{1}{e}\int_{0}^{\infty} \int_{0}^{2\pi} re^{-r^2} \; d\theta dr \qquad ((x, y) = r(\cos\theta, \sin\theta))\\ &=\frac{\pi}{e}\int_{0}^{\infty} 2re^{-r^2} \; dr\\ &=\frac{\pi}{e}. \end{align*}