Others Questions

Question Number 47770 by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{lnx}}{dx} \\$$

Commented bymaxmathsup by imad last updated on 14/Nov/18

$${you}\:{are}\:{welcome}\:{sir}. \\$$

Commented bymaxmathsup by imad last updated on 14/Nov/18

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:{changement}\:{ln}\left({x}\right)=−{t}\:{give}\:{x}=\overset{−{t}} {{e}} \\$$ $${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{e}^{−\mathrm{2}{t}} −\mathrm{1}}{−{t}}\:{e}^{−{t}} {dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}} −{e}^{−\mathrm{3}{t}} }{{t}}\:{dt}\:\:{let}\:{determine} \\$$ $${f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}} −{e}^{−\mathrm{3}{t}} }{{t}}\:{e}^{−{xt}} {dt}\:{with}\:{x}\geqslant\mathrm{0}\:{we}\:{hsve}\: \\$$ $${f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\left({e}^{−{t}} −{e}^{−\mathrm{3}{t}} \right){e}^{−{xt}} {dt}\:=−\int_{\mathrm{0}} ^{\infty} \left(\:{e}^{−\left({x}+\mathrm{1}\right){t}} −{e}^{−\left({x}+\mathrm{3}\right){t}} \right){dt}\: \\$$ $$=−\left[−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{e}^{−\left({x}+\mathrm{1}\right){t}} \:+\frac{\mathrm{1}}{{x}+\mathrm{3}}\:{e}^{−\left({x}+\mathrm{3}\right){t}} \right]_{{t}=\mathrm{0}} ^{+\infty} =−\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{{x}+\mathrm{3}}\right)\:\Rightarrow{f}\left({x}\right)=−{ln}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\right)+\lambda \\$$ $${but}\:\exists{m}>\mathrm{0}\:/\mid{f}\left({x}\right)\mid\leqslant{m}\int_{\mathrm{0}} ^{\infty} {e}^{−{xt}} {dt}\:=\frac{{m}}{{x}}\:\rightarrow\mathrm{0}\:\left({x}\rightarrow+\infty\right)\:\Rightarrow \\$$ $$\lambda={lim}_{{x}\rightarrow+\infty} \left({f}\left({x}\right)+{ln}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\right)\right)=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=−{ln}\left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{3}}\right)\:\Rightarrow{f}\left({x}\right)={ln}\left(\frac{{x}+\mathrm{3}}{{x}+\mathrm{1}}\right) \\$$ $${I}\:={f}\left(\mathrm{0}\right)={ln}\left(\mathrm{3}\right)\:\Rightarrow\:\bigstar\:{I}\:={ln}\left(\mathrm{3}\right)\bigstar\:. \\$$

Commented bytanmay.chaudhury50@gmail.com last updated on 14/Nov/18

$${thank}\:{you}\:{sir}... \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Nov/18

$${some}\:{tricks}\:{to}\:{solve}... \\$$ $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −\mathrm{1}}{{lnx}}{dx} \\$$ $$\frac{{dI}}{{da}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial\left({x}^{{a}} −\mathrm{1}\right)}{\partial{a}}×\frac{\mathrm{1}}{{lnx}}{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} ×{lnx}}{{lnx}}{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {dx} \\$$ $$=\mid\frac{{x}^{{a}+\mathrm{1}} }{{a}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{a}+\mathrm{1}} \\$$ $$\frac{{dI}}{{da}}=\frac{\mathrm{1}}{{a}+\mathrm{1}} \\$$ $${dI}=\frac{{da}}{{a}+\mathrm{1}} \\$$ $${I}={ln}\left({a}+\mathrm{1}\right)+{C} \\$$ $${put}\:\:{a}=\mathrm{0}\:\:\:{so}\:\int\frac{{x}^{{a}} −\mathrm{1}}{{lnx}}{dx}\:\:=\mathrm{0} \\$$ $${hence}\:{I}={ln}\left({a}+\mathrm{1}\right)+{C} \\$$ $$\mathrm{0}={ln}\left(\mathrm{0}+\mathrm{1}\right)+{C} \\$$ $${C}=\mathrm{0} \\$$ $${so}\:{I}={ln}\left({a}+\mathrm{1}\right) \\$$ $${so}\:{answer}\:{for}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{lnx}}{dx}={ln}\left(\mathrm{2}+\mathrm{1}\right)={ln}\mathrm{3} \\$$ $$\\$$