積分解説28

2021/01/16に Don@ld さんが出題した問題です。

https://twitter.com/atmark_donald/status/1350379594331787266?s=21

[解説]

$\begin{array}{rcl}& & {\int }_0^1\frac{t-1}{(t+1)\log t}dt\\ & =& {\int }_0^1\frac{1}{t+1}{\int }_0^1{t}^{x}dxdt\\ & =& {\int }_0^1{\int }_0^1\frac{t^x}{t+1}dtdx\\ & =& {\int }_0^1{\int }_0^1{t}^x\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{t}^{k}dtdx\\ & =& {\int }_0^1\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{\int }_0^1{t}^{x+k}dtdx\\ & =& {\int }_0^1\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{x+k+1}dx\\ & =& \frac{1}{2}{\int }_0^1(\psi (\frac{x}{2}+1)-\psi (\frac{x}{2}+\frac{1}{2}))dx\\ & =& {[\log \mathrm{\Gamma }(\frac{x}{2}+1)-\log \mathrm{\Gamma }(\frac{x}{2}+\frac{1}{2})]}_0^1\\ & =& \log \mathrm{\Gamma }\left(\frac{3}{2}\right)-\log \mathrm{\Gamma }(1)-\log \mathrm{\Gamma }(1)+\log \mathrm{\Gamma }\left(\frac{1}{2}\right)\\ & =& \log \frac{\pi }{2}\end{array}$

よって、この問題の解答は${\displaystyle \log \frac{\pi }{2}}$となります。