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Question Number 197132 by Erico last updated on 08/Sep/23

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

Commented by mnjuly1970 last updated on 09/Sep/23

$$titokaghasardool..$$

Answered by witcher3 last updated on 09/Sep/23

$$\mathrm{t}=\mathrm{sh}\left(\mathrm{x}\right)$$$${\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{x}}{\mathrm{sh}\left(\mathrm{x}\right)}\right)}^2\mathrm{ch}\left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{sh}}^2\left(\mathrm{x}\right)}\mathrm{ch}\left(\mathrm{x}\right)=\underset{(\mathrm{a},\mathrm{b})\to (0,\mathrm{\infty })}{\lim }{[-\frac{{\mathrm{x}}^2}{\mathrm{sh}\left(\mathrm{x}\right)}]}_{\mathrm{a}}^{\mathrm{b}}+{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{x}}{\mathrm{sh}\left(\mathrm{x}\right)}\mathrm{dx}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{2\mathrm{x}}{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{{4\mathrm{xe}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}$$$$=4\sum _{\mathrm{n}⩾0}{\int }_0^{\mathrm{\infty }}{\mathrm{xe}}^{-(1+2\mathrm{n})\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}⩾0}\frac{4}{{(1+2\mathrm{n})}^2}$$$$=4(\zeta \left(2\right)-\frac{1}{4}\zeta \left(2\right))=\frac{{\pi }^2}{2}$$

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