Question Number 220015 by SdC355 last updated on 04/May/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\mid\mid{J}_{\nu} \left({r}\right)\mid\mid{e}^{−{rt}} \:\mathrm{d}{r}=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{{z}_{{h}} } ^{\:{z}_{{h}+\mathrm{1}} } \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r} \\ $$$${z}_{{j}} \:\mathrm{is}\:\mathrm{point}\:\mathrm{of}\:\:{J}_{\nu}…
Question Number 220072 by hardmath last updated on 04/May/25 $$\mathrm{If}\:\:\:\mathrm{x},\mathrm{y}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mathrm{log}_{\boldsymbol{\mathrm{sinx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:+\:\mathrm{log}_{\boldsymbol{\mathrm{cosx}}} ^{\mathrm{2}} \:\left(\frac{\mathrm{sin2x}}{\mathrm{sinx}\:+\:\mathrm{cosx}}\right)\:\geqslant\:\mathrm{2} \\ $$ Answered by MrGaster last updated…
Question Number 219944 by Spillover last updated on 04/May/25 Answered by som(math1967) last updated on 04/May/25 $${Red}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}\pi×\mathrm{6}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{6}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}^{\mathrm{2}} \right) \\ $$$$\:+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×\mathrm{12}−\left(\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{6}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}^{\mathrm{2}} \right) \\…
Question Number 219945 by Spillover last updated on 04/May/25 Answered by mr W last updated on 04/May/25 Commented by mr W last updated on 04/May/25…
Question Number 220074 by Red1ight last updated on 04/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219946 by Spillover last updated on 04/May/25 Answered by Spillover last updated on 04/May/25 Answered by Spillover last updated on 04/May/25 Terms of…
Question Number 220069 by hardmath last updated on 04/May/25 $$\mathrm{Let}\:\mathrm{be}\:\:\:\left(\mathrm{H}_{\boldsymbol{\mathrm{n}}} \right)_{\boldsymbol{\mathrm{n}}\geqslant\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{H}_{\boldsymbol{\mathrm{n}}} \:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{2}\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} } \:\left(\sqrt[{\boldsymbol{\mathrm{n}}+\mathrm{1}}]{\left(\mathrm{n}+\mathrm{1}\right)!}\:−\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{n}!}\:\right)\:\mathrm{sin}\:\frac{\pi}{\mathrm{n}^{\mathrm{2}} }\:=\:? \\ $$ Answered…
Question Number 220007 by Noorzai last updated on 04/May/25 Commented by MrGaster last updated on 04/May/25 conditions are not enough and the meaning is unclear, please add or upload again if there is anything that has not been added. Answered by efronzo1 last updated on 04/May/25 $$\:{a}^{\mathrm{3}}…
Question Number 220065 by Nicholas666 last updated on 04/May/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha\in\mathbb{R}\:\:\:;\:\:\:\:\omega\in\mathbb{R}^{+} \\ $$$$\:\:\:\:\:{I}\left(\alpha\right)\:=\:\int_{−\infty} ^{\:\infty} \:\frac{{x}^{\mathrm{2}} \:\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\alpha} }\:{e}^{−{x}^{\mathrm{2}} } \:\mathrm{cos}\left(\omega{x}\right)\:{dx} \\ $$$$ \\…
Question Number 220066 by SdC355 last updated on 04/May/25 $$\mathrm{evaluate} \\ $$$$−\frac{\mathrm{csc}\left(\pi{s}\right)}{\boldsymbol{{i}}\pi}\int_{\:\boldsymbol{\mathcal{C}}} \:\left(−{t}\right)^{{s}−\mathrm{1}} {e}^{−{t}} \:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$$$−\frac{\boldsymbol{\Gamma}\left(\mathrm{1}−{s}\right)}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{\:\boldsymbol{\mathcal{C}}} \:\frac{\left(−{t}\right)^{{s}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}\:\mathrm{d}{t}\:,\:\mathrm{path}\:\boldsymbol{\mathcal{C}};\left(−\infty,\infty\right) \\ $$ Terms of Service…