证明: $$\bex \int_0^\infty \frac{\rd x}{1+x^4}=\int_0^\infty \frac{x^2}{1+x^4}\rd x=\frac{\pi}{2\sqrt{2}}. \eex$$ (北京航空航天大学)
证明: $$\beex\bea I&\equiv \int_0^\infty \frac{\rd x}{1+x^4} =\int_0^\infty \frac{1}{1+\sex{\frac{1}{t}}^4}\cdot \frac{1}{t^2}\rd t\quad\sex{t=\frac{1}{x}}\\ & =\int_0^\infty \frac{t^2}{1+t^4}\rd t =\frac{1}{2}\int_0^\infty \frac{1+t^2}{1+t^4}\rd t =\frac{1}{2}\int_0^\infty \frac{\frac{1}{t^2}+1}{\frac{1}{t^2}+t^2}\rd t\\ &=\frac{1}{2}\int_0^\infty \frac{1}{\sex{t-\frac{1}{t}}^2+2}\rd \sex{t-\frac{1}{t}} =\frac{1}{2}\int_{-\infty}^{+\infty} \frac{1}{s^2+2}\rd s\quad\sex{s=t-\frac{1}{t}}\\ &=\frac{1}{\sqrt{2}}\int_0^\infty \frac{1}{1+\sex{\frac{s}{\sqrt{2}}}^2}\rd \frac{s}{\sqrt{2}} =\frac{1}{\sqrt{2}}\cdot \frac{\pi}{2} =\frac{\pi}{2\sqrt{2}}. \eea\eeex$$
