積分解説03

2020/11/01に出題した問題です。
https://twitter.com/sounansya_29/status/1322861591889547267?s=21

[解説]
$\begin{array}{rcl}& & {\int }_0^{\frac{\pi }{2}}{e}^{-\frac{\pi }{2}\tan x}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-\frac{\pi }{2}t}}{1+t^2}dt& (t=\tan x)\\ & =& {\int }_0^{\mathrm{\infty }}{e}^{-\frac{\pi }{2}t}{\int }_0^{\mathrm{\infty }}{e}^{-ut}\sin ududt\\ & =& {\int }_0^{\mathrm{\infty }}\sin u{\int }_0^{\mathrm{\infty }}{e}^{-(u+\frac{\pi }{2})t}dtdu\\ & =& {\int }_0^{\mathrm{\infty }}\sin u{[-\frac{1}{u+\frac{\pi }{2}}{e}^{-(u+\frac{\pi }{2})t}]}_0^{\mathrm{\infty }}du\\ & =& {\int }_0^{\mathrm{\infty }}\frac{\sin u}{u+\frac{\pi }{2}}du\\ & =& -{\int }_{\frac{\pi }{2}}^{\mathrm{\infty }}\frac{\cos v}{v}dv& (v=u+\frac{\pi }{2})\\ & =& \text{Ci}\left(\frac{\pi }{2}\right)\end{array}$
よって、この問題の解答は${\displaystyle \text{Ci}\left(\frac{\pi }{2}\right)}$となります。