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Prove-that-inequality-0-1-ln-1-x-2-1-x-2-dx-lt-0-1-x-ln-1-x-2-1-x-2-dx-1-3-

Question Number 220544 by Nicholas666 last updated on 15/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{inequality}}; \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx}\:<\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\:\mathrm{ln}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} \:}\:{dx}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\: \\ $$$$ \\…

0-ln-z-2-1-z-2-1-dz-I-I-t-0-ln-z-2-1-z-2-1-e-zt-dz-I-t-0-z-ln-z-2-1-z-2-1-e-zt-dz-I-t-0-z-2-ln-z-2-1-ln-z-2-1-ln-

Question Number 220562 by SdC355 last updated on 15/May/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{z}\centerdot\mathrm{ln}\left({z}^{\mathrm{2}}…

Find-n-1-1-n-1-n-3-n-1-3-2n-1-2-

Question Number 220540 by hardmath last updated on 14/May/25 $$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{n}^{\mathrm{3}} \centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)^{\mathrm{3}} \centerdot\left(\mathrm{2n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by cadmon98 last updated on 16/May/25…

each-J-z-Y-z-are-linear-independent-W-Ronskian-J-z-Y-z-determinant-J-z-Y-z-J-z-Y-z-J-1-z-Y-z-J-z-Y-1-z-J-1-

Question Number 220502 by SdC355 last updated on 14/May/25 $$\mathrm{each}\:{J}_{\nu} \left({z}\right),{Y}_{\nu} \left({z}\right)\:\mathrm{are}\:\mathrm{linear}\:\mathrm{independent}….?? \\ $$$${W}_{\mathrm{Ronskian}} \left\{{J}_{\nu} ^{\:} \left({z}\right),{Y}_{\nu} \left({z}\right)\right\}=\begin{vmatrix}{{J}_{\nu} \left({z}\right)}&{\:{Y}_{\nu} \left({z}\right)}\\{{J}_{\nu} '\left({z}\right)}&{{Y}_{\nu} '\left({z}\right)}\end{vmatrix} \\ $$$$={J}_{\nu} ^{\left(\mathrm{1}\right)}…