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Integration Questions

Question Number 134302 by bramlexs22 last updated on 02/Mar/21

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\mathrm{dx} \\$$

Answered by Dwaipayan Shikari last updated on 02/Mar/21

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\:\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)} \\$$ $${I}'\left({a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\pi}\:\frac{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\Gamma'\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)} \\$$ $${I}'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{logx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{\sqrt{\pi}}{\mathrm{4}}.\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\psi\left(\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{4}}\left(−\gamma−\mathrm{2}{log}\left(\mathrm{2}\right)+\gamma\right) \\$$ $$=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\$$

Commented byDwaipayan Shikari last updated on 02/Mar/21

$${Another}\:{way}\:{to}\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\theta\right){d}\theta=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\$$

Commented bymnjuly1970 last updated on 02/Mar/21

$$\:{yes}\:{sir}.. \\$$ $$\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\$$ $$\:\:\:\:\:\boldsymbol{\phi}\overset{\langle{sin}\left({x}\right)={t}\rangle} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right)\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\$$

Answered by bramlexs22 last updated on 02/Mar/21

$$\mathrm{let}\:{x}=\mathrm{sin}\:\beta\: \\$$ $$\mathrm{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\mathrm{ln}\:\left(\mathrm{sin}\:\beta\right)\:\mathrm{d}\beta=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\mathrm{ln}\:\left(\mathrm{cos}\:\beta\right)\mathrm{d}\beta \\$$ $$=\:−\frac{\pi}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}\right) \\$$