Question Number 35832 by abdo mathsup 649 cc last updated on 24/May/18
$${find}\:{the}\:{value}\:{of}\:\:{f}\left(\lambda\right)\:=\:\int_{−{a}} ^{{a}} \:\:\:\frac{{dx}}{\left(\lambda\:+_{} {x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\lambda\in{R}\:. \\ $$
Commented by abdo.msup.com last updated on 25/May/18
$${case}\:\mathrm{1}\:{if}\lambda>\mathrm{0}\:{let}\:{use}\:{the}\:{changement} \\ $$$${x}=\sqrt{\lambda}\:{tan}\left({t}\right) \\ $$$${f}\left(\lambda\right)\:=\:\int_{−{arctan}\left(\frac{{a}}{\lambda}\right)} ^{{arctan}\left(\frac{{a}}{\lambda}\right)} \:\:\:\frac{\sqrt{\lambda}}{\lambda^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \left({t}\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\lambda}\int_{−{arctan}\left(\frac{{a}}{\:\sqrt{\lambda}}\right)} ^{{arctan}\left(\:\frac{{a}}{\:\sqrt{\lambda}}\right)\:} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt} \\ $$$$=\:\frac{\mathrm{2}}{\lambda}\:\int_{\mathrm{0}} ^{{arctan}\left(\frac{{a}}{\:\sqrt{\lambda}}\right)} \:\:{cos}\left({t}\right)\:{dt} \\ $$$$=\frac{\mathrm{2}}{\lambda}\:\left[\:{sin}\left({t}\right)\right]_{\mathrm{0}} ^{{arctan}\left(\frac{{a}}{\:\sqrt{\lambda}}\right)} \\ $$$$=\:\frac{\mathrm{2}}{\lambda}\:{sin}\left({arctan}\left(\frac{{a}}{\:\sqrt{\lambda}}\right)\right) \\ $$$${but}\:{we}\:{have}\:{sin}\left({arctanx}\right)\:=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${so}\:{sin}\left({arctan}\left(\frac{{a}}{\:\sqrt{\lambda}}\right)\right)\:=\:\frac{{a}}{\:\sqrt{\lambda}\:\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\lambda}}} \\ $$$$=\:\frac{{a}}{\:\sqrt{\lambda\:+{a}^{\mathrm{2}} }}\:\Rightarrow{f}\left(\lambda\right)\:=\:\frac{\mathrm{2}{a}}{\lambda\sqrt{\lambda+{a}^{\mathrm{2}} }} \\ $$
Commented by prof Abdo imad last updated on 25/May/18
$${case}\:\mathrm{2}\:{if}\:\:\lambda<\mathrm{0}\:\Rightarrow\:−\lambda\:>\mathrm{0}\:{let}\:{put}\:\alpha\:=−\lambda \\ $$$${f}\left(\lambda\right)\:=\:\int_{−{a}} ^{{a}} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:−\alpha\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:.{changement}\:{x}=\sqrt{\alpha}{ch}\left({t}\right) \\ $$$${give}\: \\ $$$${f}\left(\lambda\right)\:=\:\int_{{argch}\left(\:\frac{−{a}}{\:\sqrt{\alpha}}\right)} ^{{argch}\left(\:\frac{{a}}{\:\sqrt{\alpha}}\right)} \:\:\:\:\:\frac{\sqrt{\alpha}\:\:{sh}\left({t}\right)}{\alpha^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\:{ch}^{\mathrm{2}} \left({t}\right)−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt} \\ $$$$=\:\frac{\mathrm{1}}{\alpha}\int_{{argch}\left(\frac{−{a}}{\:\sqrt{\alpha}}\right)} ^{{argch}\left(\:\frac{{a}}{\:\sqrt{\alpha}}\right)} \:\:\:\:\frac{{sh}\left({t}\right)}{{sh}^{\mathrm{3}} \left({t}\right)}{dt}\: \\ $$$$=\:\frac{\mathrm{2}}{\alpha}\:\int_{{argch}\left(\:\frac{−{a}}{\:\sqrt{\alpha}}\right)} ^{{argch}\left(\frac{{a}}{\:\sqrt{\alpha}}\right)} \:\:\:\:\:\frac{{dt}}{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2}}{\alpha}\:\int_{{ln}\left(\:−\frac{{a}}{\:\sqrt{\kappa}}\:\:+\sqrt{\:\frac{{a}^{\mathrm{2}} }{\alpha}\:−\mathrm{1}}\right)} ^{{ln}\left(\:\frac{{a}}{\:\sqrt{\alpha}}\:+\sqrt{\:\frac{{a}^{\mathrm{2}} }{\alpha}−\mathrm{1}}\right.} \:\:\:\:\:\:\:\frac{{dt}}{\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{4}}{\alpha}\:\int_{{ln}\left(−\frac{{a}}{\:\sqrt{\alpha}}\:+\sqrt{\:\frac{{a}^{\mathrm{2}} }{\alpha}−\mathrm{1}}\right)} ^{{ln}\left(\:\frac{{a}}{\:\sqrt{\alpha}}\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\alpha}−\mathrm{1}}\right)} \:\:\:\:\:\frac{{dt}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:−\mathrm{2}}\:… \\ $$$${be}\:{continued}… \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 26/May/18
$${let}\:{find}\:{K}=\int\:\:\:\:\:\:\:\frac{{dt}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:−\mathrm{2}}\:\:{chamgement} \\ $$$${e}^{\mathrm{2}{t}} \:={u}\:{give}\:\:{K}\:=\:\int\:\:\frac{\mathrm{1}}{{u}\:+{u}^{−\mathrm{1}} \:−\mathrm{2}}\:\frac{{du}}{\mathrm{2}{u}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}\:−\mathrm{2}{u}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}\left({u}−\mathrm{1}\right)}\:\:=\frac{−\mathrm{1}}{\mathrm{2}\left({e}^{\mathrm{2}{t}} \:−\mathrm{1}\right)}\:+{c}\:{so} \\ $$$${f}\left(\lambda\right)\:=\:−\frac{\mathrm{4}}{\mathrm{2}\alpha}\:\left[\:\:\frac{\mathrm{1}}{{e}^{\mathrm{2}{t}} −\mathrm{1}}\right]_{{ln}\left(\frac{−{a}}{\:\sqrt{\alpha}}\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\alpha}−\mathrm{1}}\right.} ^{{ln}\left(\:\frac{{a}}{\:\sqrt{\alpha}}\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\alpha}\:−\mathrm{1}}\right.} \\ $$$$=\:\frac{−\mathrm{2}}{\alpha}\left\{\:\:\:\:\:\frac{\mathrm{1}}{\left(\frac{{a}}{\:\sqrt{\alpha}}\:+\sqrt{\frac{{a}^{\mathrm{2}} }{\alpha}−\mathrm{1}}\right)}\:−\:\frac{\mathrm{1}}{\left(−\frac{{a}}{\:\sqrt{\alpha}}+\sqrt{\frac{{a}^{\mathrm{2}} }{\alpha}\:−\mathrm{1}}\right)}\right. \\ $$$$=\:\frac{\mathrm{2}}{\lambda}\left\{\:\:\:\:\:\frac{\mathrm{1}}{\left(\:\frac{{a}}{\:\sqrt{−\lambda}}\:+\sqrt{−\frac{{a}^{\mathrm{2}} }{\lambda}−\mathrm{1}}\right)}−\frac{\mathrm{1}}{\left(\frac{−{a}}{\:\sqrt{−\lambda}}+\sqrt{−\frac{{a}^{\mathrm{2}} }{\lambda}−\mathrm{1}}\right)}\right. \\ $$$${with}\:\lambda<\mathrm{0}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/May/18
$${x}=\sqrt{\lambda}\:{tan}\theta\:\:\:{dx}=\sqrt{\lambda}\:{sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$${I}=\int\frac{\sqrt{\lambda}\:{sec}^{\mathrm{2}} \theta\:{d}\theta}{\left\{\lambda\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\int\frac{\mathrm{1}}{\lambda}\:×\frac{{sec}^{\mathrm{2}} \theta}{{sec}^{\mathrm{3}} \theta}×\frac{{d}\theta}{} \\ $$$$=\frac{\mathrm{1}}{\lambda}\int{cos}\theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\lambda}{sin}\theta\:\:{now}\:{putting}\:{the}\:{limit} \\ $$$$=\frac{\mathrm{1}}{\lambda}×\mid\frac{{x}}{\left(\lambda+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:}\mid_{−{a}} ^{{a}} \\ $$$$=\frac{\mathrm{1}}{\lambda}×\frac{\mathrm{2}{a}}{\left(\lambda+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$
Commented by abdo.msup.com last updated on 25/May/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{Tanmay} \\ $$$${but}\:{you}\:{must}\:{study}\:{the}\:{case}\:\lambda<\mathrm{0}…. \\ $$