積分２

解説
$\arctan x=\pi y$とすると,上の積分は
$${\int }_0^{\frac{1}{4}}({\displaystyle \frac{\pi }{2}}{e}^{\frac{1}{2}\tan \pi y}(1+{\tan }^2\pi y)\sin 2\pi y+2\pi e^{\frac{1}{2}\tan \pi y}\cos 2\pi y)dy$$
となります.
ここで,
$${\mathrm{I}}_1={\int }_0^{\frac{1}{4}}2\pi e^{\frac{1}{2}\tan \pi y}\cos 2\pi ydy,{\mathrm{I}}_2={\int }_0^{\frac{1}{4}}{\displaystyle \frac{\pi }{2}}{e}^{\frac{1}{2}\tan \pi y}(1+{\tan }^2\pi y)\sin 2\pi ydy$$
とおくと,求める積分は${\mathrm{I}}_1+{\mathrm{I}}_2$となります.
$$\begin{array}{rcl}{\mathrm{I}}_1& =& {\int }_0^{\frac{1}{4}}2\pi e^{\frac{1}{2}\tan \pi y}\cos 2\pi ydy\\ & =& [e^{\frac{1}{2}\tan \pi y}\sin 2y{]}_0^{\frac{1}{4}}-{\int }_0^{\frac{1}{4}}{\displaystyle \frac{\pi }{2}}{e}^{\frac{1}{2}\tan \pi y}(1+{\tan }^2\pi y)\sin 2\pi ydy\\ & =& \sqrt{e}-{\mathrm{I}}_2\\ {\mathrm{I}}_1+{\mathrm{I}}_2& =& \sqrt{e}\end{array}$$
よって積分の答えは$\sqrt{e}$となります.