Question Number 222275 by Nicholas666 last updated on 21/Jun/25 $$ \\ $$$$\:\:\mathrm{Prove}\:;\:\int_{−\pi} ^{\:\pi} \:\frac{{z}\:\mathrm{sin}\left({z}\right)\:}{\left(\mathrm{1}\:+\:{z}\:+\:\sqrt{\mathrm{1}\:+\:{z}^{\mathrm{2}} }\right)\sqrt{\mathrm{3}\:+\:\mathrm{sin}^{\mathrm{2}} \left({z}\right)}}\:{dz}\:=\:\zeta\left(\mathrm{2}\right)\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Answered by MrGaster last updated…
Question Number 222224 by MrGaster last updated on 20/Jun/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\mathrm{cos}\left(\mathrm{ln}\:{x}\right){dx}=\mathrm{1}−\frac{\pi}{\mathrm{2}}\mathrm{cosh}\frac{\pi}{\mathrm{2}} \\ $$ Commented by Nicholas666 last updated on 21/Jun/25 $$\:\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{statment}\:\mathrm{is}\:\mathrm{wrong},\: \\ $$$$\mathrm{no}\:\mathrm{1}−\frac{\pi}{\mathrm{2}}\:\mathrm{cosh}\:\frac{\pi}{\mathrm{2}}\:\mathrm{but}\:\mathrm{1}\:−\frac{\pi}{\mathrm{2}}\mathrm{chot}\:\left(\frac{\pi}{\mathrm{2}}\right)…
Question Number 222217 by MrGaster last updated on 20/Jun/25 $$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\mathrm{Sol}:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right){dxdy}=\mathrm{Re}\left(\left(\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}\right)\left(\int_{\mathrm{0}}…
Question Number 222218 by Nicholas666 last updated on 20/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{ln}\:\mathrm{sin}\left({x}\right)}{{x}}\right)\:{dx} \\ $$$$ \\ $$ Answered by Nicholas666 last updated on…
Question Number 222245 by Nicholas666 last updated on 20/Jun/25 $$ \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)\:\mathrm{cos}^{\mathrm{2}} \left({x}\right)}}{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}\right)\centerdot\mathrm{ln}\left(\frac{\mathrm{1}\:+\:\mathrm{sin}\left({x}\right)}{\mathrm{1}\:+\:\mathrm{cos}\left({x}\right)}\right)}{\:\sqrt{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \left({x}\right)\:−\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}}\:\:\mathrm{d}{x}\:\:\:\:\:\: \\ $$$$ \\ $$ Answered…
Question Number 222175 by klipto last updated on 19/Jun/25 $$\boldsymbol{\mathrm{solve}} \\ $$$$\left(\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{y}}} −\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{cosx}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$ Answered by som(math1967) last updated on 20/Jun/25…
Question Number 222127 by Nicholas666 last updated on 18/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left({e}^{\frac{\mathrm{1}}{\pi}\:\mathrm{tan}^{−\mathrm{1}} \:{u}} \right)\right)\right)\:}{{u}^{\mathrm{2}} \:+\:\mathrm{2}\pi{u}\:+\:\mathrm{2}\pi^{\mathrm{2}} }\:{du} \\ $$$$ \\ $$ Terms of…
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Question Number 222064 by MathematicalUser2357 last updated on 16/Jun/25 Commented by MathematicalUser2357 last updated on 19/Jun/25 $$\mathrm{Mister}\:\mathrm{MathML} \\ $$ Commented by Nicholas666 last updated on…
Question Number 222019 by Nicholas666 last updated on 15/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{ln}\left(\mathrm{sin}\left({x}\right)\right.}{{x}}\right)\:{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…