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Question Number 197113 by Erico last updated on 08/Sep/23

$$\mathrm{Prove}\mathrm{that}$$$${\underset{0}{\int }}^{+\mathrm{\infty }}\left(\frac{ln(t+\sqrt{1+t^2})}{t}\right)dt=\frac{{\pi }^2}{2}$$

Answered by witcher3 last updated on 08/Sep/23

$$\ln (\mathrm{t}+\sqrt{1+{\mathrm{t}}^2})>2,\mathrm{\forall }\mathrm{x}⩾\mathrm{e}$$$${\int }_2^{\mathrm{\infty }}\frac{\ln (\mathrm{t}+\sqrt{{\mathrm{t}}^2+1})}{\mathrm{t}}⩾{\int }_2^{\mathrm{\infty }}\frac{2}{\mathrm{t}}\mathrm{dt}=\mathrm{\infty }\mathrm{integral}\mathrm{didnt}\mathrm{cv}$$

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