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Question-62263

Question Number 62263 by Tawa1 last updated on 18/Jun/19 Answered by behi83417@gmail.com last updated on 19/Jun/19 $$\mathrm{x}+\sqrt{\mathrm{x}−\mathrm{2}}=\mathrm{a}\Rightarrow\frac{\mathrm{2}}{\mathrm{a}}=\frac{\mathrm{1}}{\mathrm{x}+\sqrt{\mathrm{x}−\mathrm{2}}}=\frac{\mathrm{x}−\sqrt{\mathrm{x}−\mathrm{2}}}{\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{\mathrm{a}}+\sqrt{\mathrm{a}}=\mathrm{3}\Rightarrow\mathrm{a}\sqrt{\mathrm{a}}−\mathrm{3a}+\mathrm{2}=\mathrm{0} \\ $$$$\sqrt{\mathrm{a}}=\mathrm{t}\Rightarrow\mathrm{t}^{\mathrm{3}} −\mathrm{3t}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{t}−\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}}…

find-the-value-of-I-0-e-t-sint-t-dt-and-J-0-e-t-cos-t-t-dt-study-first-the-convergence-

Question Number 62262 by maxmathsup by imad last updated on 18/Jun/19 $${find}\:{the}\:{value}\:{of}\: \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} {sint}}{\:\sqrt{{t}}}{dt}\:\:{and}\:{J}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} {cos}\left({t}\right)}{\:\sqrt{{t}}}{dt}\:\:,{study}\:{first}\:{the}\:{convergence}. \\ $$ Commented by maxmathsup…