Integration Questions

Question Number 123159 by Lordose last updated on 23/Nov/20

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{log}^{\mathrm{2}} \left(\mathrm{tan}\left(\mathrm{x}\right)\right)\mathrm{dx} \\$$

Answered by mnjuly1970 last updated on 23/Nov/20

$${ans}:=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\$$

Commented by Lordose last updated on 23/Nov/20

$$\mathrm{It}'\mathrm{s}\:\mathrm{correct}\:\mathrm{sir} \\$$

Commented by mnjuly1970 last updated on 23/Nov/20

$${thank}\:{you} \\$$$${solution} \\$$$${I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {log}^{\mathrm{2}} \left({tan}\left({x}\right)\right){dx}\overset{{tan}\left({x}\right)={y}} {=} \\$$$$\:\:=\int_{\mathrm{0}} ^{\:\infty} {log}^{\mathrm{2}} \left({y}\right)\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }\overset{{y}^{\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\infty} \:\frac{{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} {log}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}+{t}}\:{dt} \\$$$$\\$$$$\:{I}\left({a}\right)\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\frac{−\mathrm{1}}{\mathrm{2}}+{a}} }{\mathrm{1}+{t}}{dt} \\$$$$\:\:{goal}\:::\:\:{I}=\frac{\mathrm{1}}{\mathrm{8}}{I}\:''\left(\mathrm{0}\right) \\$$$$\:\:\:\:{but}\::\:\:{I}\left({a}\right)=\beta\left(\frac{\mathrm{1}}{\mathrm{2}}+{a},\frac{\mathrm{1}}{\mathrm{2}}−{a}\right)\:\:\:\:\:\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{a}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−{a}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{reflection}\:{formula}} {\overset{{euler}} {=}}\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}+\pi{a}\right)}=\frac{\pi}{{cos}\left(\pi{a}\right)} \\$$$$\:{I}\:'\left({a}\right)=\frac{\pi^{\mathrm{2}} {sin}\left(\pi{a}\right)}{{cos}^{\mathrm{2}} \left(\pi{a}\right)} \\$$$$\:\:{I}\:''\left({a}\right)=\pi^{\mathrm{2}} \left(\frac{\pi{cos}^{\mathrm{3}} \left(\pi{a}\right)+\mathrm{2}\pi{sin}^{\mathrm{2}} \left(\pi{a}\right){cos}\left(\pi{a}\right)}{{cos}^{\mathrm{4}} \left(\pi{a}\right)}\right)\mid_{{a}=\mathrm{0}} \\$$$$\:\:\:{I}\:''\left(\mathrm{0}\right)=\pi^{\mathrm{3}} \Rightarrow\:{I}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}}\:\:\checkmark\checkmark \\$$$$\\$$$$\\$$$$\\$$

Commented by Dwaipayan Shikari last updated on 23/Nov/20

$${Great}\:{sir}!\:{i}\:{haven}'{t}\:{thought}\:{of}\:{this} \\$$

Commented by mnjuly1970 last updated on 23/Nov/20

$$\:\:\:{thank}\:{you}\:{so}\:{much}\:{sir}\:{payan} \\$$$$\:\:{to}\:{me}\:{your}\:{solution}\:{is}\:{very}\:{very}\: \\$$$${nice}\:{and}\:{unique}... \\$$

Answered by Dwaipayan Shikari last updated on 23/Nov/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}^{\mathrm{2}} \left({tanx}\right){dx} \\$$$$=\int_{\mathrm{0}} ^{\infty} \frac{{log}^{\mathrm{2}} {t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\: \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} {log}^{\mathrm{2}} {t}\:\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} {dt}\:\:\:\:\:\:\:\:{logt}={u}\Rightarrow\frac{\mathrm{1}}{{t}}=\frac{{du}}{{dt}} \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{−\infty} ^{\infty} {u}^{\mathrm{2}} {e}^{\mathrm{2}{nu}+{u}} {dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}\left(\mathrm{2}{n}+\mathrm{1}\right)=\Lambda \\$$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \Lambda^{\mathrm{2}} {e}^{−\Lambda} {d}\Lambda \\$$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\Gamma\left(\mathrm{3}\right) \\$$$$=\mathrm{4}\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{4}.\frac{\pi^{\mathrm{3}} }{\mathrm{32}}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\$$

Commented by mnjuly1970 last updated on 23/Nov/20

$${very}\:{good}\:... \\$$

Commented by Lordose last updated on 23/Nov/20

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{from}\:\mathrm{4th}\:\mathrm{line} \\$$

Commented by Dwaipayan Shikari last updated on 24/Nov/20

$$\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{−\infty} ^{\infty} \frac{\Lambda^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)}{e}^{\Lambda} {d}\Lambda\:\:\:\:\:\:{u}\left(\mathrm{2}{n}+\mathrm{1}\right)=\Lambda \\$$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\infty} ^{−\infty} −\Phi^{\mathrm{2}} {e}^{−\Phi} \:{d}\Phi\:\:\:\:\:\:\:\:\:\Lambda=−\Phi \\$$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \Phi^{\mathrm{2}} {e}^{−\Phi} {d}\Phi\:\:\:=\mathrm{2}\:.\Gamma\left(\mathrm{3}\right)\frac{\pi^{\mathrm{3}} }{\mathrm{32}}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\$$

Answered by mathmax by abdo last updated on 23/Nov/20

$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{tanx}\right)\mathrm{dx}\:\mathrm{changementtanx}=\mathrm{t}\:\mathrm{give} \\$$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:+\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\left(\rightarrow\mathrm{t}=\frac{\mathrm{1}}{\mathrm{x}}\right) \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}\left(−\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} }\right)\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)\left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} \right)\mathrm{dx}\:=\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{dx} \\$$$$\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}\:} \mathrm{x}^{\mathrm{2n}} \:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{dx}\:=\left[\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\frac{\mathrm{2lnx}}{\mathrm{x}}\mathrm{dx} \\$$$$=−\frac{\mathrm{2}}{\mathrm{2n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\:=−\frac{\mathrm{2}}{\mathrm{2n}+\mathrm{1}}\left\{\left[\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\mathrm{lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}} }{\left(\mathrm{2n}+\mathrm{1}\right)}\mathrm{dx}\right\} \\$$$$=\frac{\mathrm{2}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\:\mathrm{A}\:=\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }\:\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{tbe}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this} \\$$$$\mathrm{serie}\:\mathrm{by}\:\mathrm{fourier}.... \\$$