微分方程式で積分を解く

$$\begin{array}{rl}I& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{(x^2+1{)}^2}\\ & ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^{2}xdx\\ & =\frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1+\cos 2x)dx\\ & =\frac{1}{2}{[x+2\sin 2x]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\\ & =\frac{\pi }{2}\end{array}$$

全ての実数$x$で$(x^2+1{)}^2\ne 0$かつ(分母の次数)$\ge$(分子の次数)　なので広義積分は収束し、被積分関数が奇関数なので積分値は$0$

$$\begin{array}{rl}I& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\cos }^{2}x}{(x^2+1{)}^2}dx\\ & =\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{(x^2+1{)}^2}+\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos 2x}{(x^2+1{)}^2}dx\\ & =\frac{\pi }{4}+\frac{1}{2}J\end{array}$$

$$J={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos 2x}{(x^2+1{)}^2}$$

$$f(t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos tx}{(x^2+1{)}^2}dx$$
$$f^{\prime }(t)=-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x\sin tx}{(x^2+1{)}^2}dx$$
$$f^{″}(t)=-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2\cos tx}{(x^2+1{)}^2}dx$$

$$\begin{array}{rl}f(t)-f^{″}(t)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos tx}{x^2+1}dx\\ & =\pi e^{-t}\end{array}$$
$$\begin{array}{rl}f(0)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{(x^2+1{)}^2}\\ & =\frac{\pi }{2}\end{array}$$
$$\begin{array}{rl}{f}^{\prime }(0)& =-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x}{(x^2+1{)}^2}dx\\ & =0\end{array}$$

$$f(t)-f^{″}(t)=\pi e^{-t}$$
$$f(0)=\frac{\pi }{2}$$
$$f^{\prime }(0)=0$$
微分方程式を解くと
$$f(t)=\frac{\pi }{2}{e}^{-t}(t+1)$$

$$\begin{array}{rl}J& =f(2)\\ & =\frac{3\pi }{2e^2}\end{array}$$

$$\begin{array}{rl}I& =\frac{\pi }{4}+\frac{1}{2}\frac{3\pi }{3e^2}\\ & =\frac{\pi }{4}(1+\frac{3}{e^2})\end{array}$$