Question Number 197247 by Erico last updated on 11/Sep/23
$$\mathrm{calcul}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{{ln}\left({cht}\right)}{{sh}\left({t}\right)}{dt} \\ $$
Answered by witcher3 last updated on 11/Sep/23
$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{ch}\left(\mathrm{t}\right)\right)\mathrm{sh}\left(\mathrm{t}\right)}{\mathrm{1}−\mathrm{ch}^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$$$\mathrm{ch}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$\Leftrightarrow\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{y}\right)}{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\mathrm{dy},\mathrm{y}=\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{y}\right)}{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }=−\Sigma\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}^{\mathrm{2k}} \mathrm{ln}\left(\mathrm{y}\right)=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by Erico last updated on 12/Sep/23
$$\mathrm{thank} \\ $$