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Question Number 197247 by Erico last updated on 11/Sep/23

$$\mathrm{calcul}{\underset{0}{\int }}^{+\mathrm{\infty }}\frac{ln\left(cht\right)}{sh\left(t\right)}dt$$

Answered by witcher3 last updated on 11/Sep/23

$$={\int }_0^{\mathrm{\infty }}\frac{\ln \left(\mathrm{ch}\left(\mathrm{t}\right)\right)\mathrm{sh}\left(\mathrm{t}\right)}{1-{\mathrm{ch}}^2\left(\mathrm{x}\right)}$$$$\mathrm{ch}\left(\mathrm{x}\right)=\mathrm{y}$$$$\iff {\int }_1^{\mathrm{\infty }}\frac{\ln \left(\mathrm{y}\right)}{1-{\mathrm{y}}^2}\mathrm{dy},\mathrm{y}=\frac{1}{\mathrm{z}}$$$$=-{\int }_0^1\frac{\ln \left(\mathrm{y}\right)}{1-{\mathrm{y}}^2}=-\mathrm{\Sigma }{\int }_0^1{\mathrm{y}}^{2\mathrm{k}}\ln \left(\mathrm{y}\right)=\sum _{\mathrm{k}⩾0}\frac{1}{{(2\mathrm{k}+1)}^2}$$$$=\frac{{\pi }^2}{8}$$

Commented by Erico last updated on 12/Sep/23

$$\mathrm{thank}$$

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