Permutation and Combination Questions

Question Number 108262 by 675480065 last updated on 15/Aug/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{y}^{\mathrm{y}} \right)^{\left(\mathrm{y}^{\mathrm{y}} \right)^{\left(\mathrm{y}^{\mathrm{y}} \right)} } \mathrm{dy}=? \\$$ $$\mathrm{please}\:\mathrm{help} \\$$

Answered by mathdave last updated on 15/Aug/20

$${answer}\:{is}\:\frac{\pi}{\mathrm{12}} \\$$

Commented by675480065 last updated on 15/Aug/20

$$\mathrm{solution}\:\mathrm{pls} \\$$

Answered by mathdave last updated on 16/Aug/20

$${solution} \\$$ $${let}\:{x}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({y}^{{y}} \right)^{\left({y}^{{y}} \right)^{\left({y}^{{y}} \right)^{.^{.^{..^{\bullet} } } } } } {dy}\:\:\:\:{therefor}\:\:{x}={y}^{{yx}\:} {let}\:\:\:{t}={yx} \\$$ $$\frac{{t}}{{y}}={y}^{{y}×\frac{{t}}{{y}}} ={y}^{{t}} \:{and}\:{x}=\frac{{t}}{{y}} \\$$ $${t}={y}^{{t}} .{y}={y}^{{t}+\mathrm{1}} \:\:\:\:{den}\:\:{y}={t}^{\frac{\mathrm{1}}{\mathrm{1}+{t}}} ..............\left(\mathrm{1}\right) \\$$ $${but}\:{x}=\frac{{t}}{{y}}={t}^{\frac{{t}}{\frac{\mathrm{1}}{\mathrm{1}+{t}}}} ={t}.{t}^{−\frac{\mathrm{1}}{\mathrm{1}+{t}}} \:\:\:\:\:\because\:\:{x}={t}^{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)} ..........\left(\mathrm{2}\right) \\$$ $${but}\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} {xdy}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)} \frac{{d}}{{dt}}\left({t}^{\frac{\mathrm{1}}{\mathrm{1}+{t}}} \right){dt} \\$$ $${since}\:\:{y}^{\frac{\mathrm{1}}{\mathrm{1}+{t}}} \:\:{logging}\:{both}\:{side}\:\:{and}\:{differentiate} \\$$ $${using}\:{product}\:{formular} \\$$ $$\frac{{d}}{{dt}}\left(\mathrm{ln}{y}\right)=\frac{\mathrm{1}}{\mathrm{1}+{t}}.\frac{{d}}{{dt}}\left(\mathrm{ln}{t}\right)+\mathrm{ln}{t}\frac{{d}}{{dt}}\left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right) \\$$ $$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dt}}=\left[\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}\right)}−\frac{\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right]={y}\left[\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}\right)}−\frac{\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right] \\$$ $$\frac{{dy}}{{dt}}={t}^{\frac{\mathrm{1}}{\mathrm{1}+{t}}} \left[\frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}\right)}−\frac{\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right]\:\:\:\:{then}\:{putting}\:\:{back} \\$$ $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{{t}}{\mathrm{1}+{t}}} .{t}^{\frac{\mathrm{1}}{\mathrm{1}+{t}}} .\frac{\mathrm{1}}{{t}}\left[\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{{t}\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right]{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}.{t}^{−\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{{t}\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right]{dt} \\$$ $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\$$ $${I}=\mathrm{ln}\left(\mathrm{1}+{t}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\mathrm{ln2}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({t}+\mathrm{1}−\mathrm{1}\right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\mathrm{ln}{tdt} \\$$ $${I}_{{AB}} =\mathrm{ln2}−\left[\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+{t}\right)\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}\right] \\$$ $${by}\:{solving}\:{B}\:{using}\:{I}.{B}.{p} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{t}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=−\frac{\mathrm{ln}{t}}{\mathrm{1}+{t}}\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}}{dt}=\mathrm{ln}{t}\mid_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{ln}\left(\mathrm{1}+{t}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\$$ $$\because\:{I}_{{B}} =−\mathrm{ln2}\:\:\:\:\:{and}\:{solving}\:{I}_{{A}} \:\:{using}\:{I}.{B}.{P} \\$$ $${I}_{{A}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}}{dt}=\mathrm{ln}\left(\mathrm{1}+{t}\right)\mathrm{ln}{t}\mid_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt} \\$$ $${using}\:{maclaurin}'{s}\:{series}\: \\$$ $$\mathrm{ln}\left(\mathrm{1}+{t}\right)=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{{n}} }{{n}}=−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {t}^{{n}} }{{n}}.\frac{\mathrm{1}}{{t}}{dt} \\$$ $${I}_{{A}=} =−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}−\mathrm{1}} {dt}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\left[\frac{{t}^{{n}−\mathrm{1}+\mathrm{1}} }{{n}−\mathrm{1}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$ $${I}_{{A}} =−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=−\eta\left(\mathrm{2}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\$$ $${but}\:{I}=\mathrm{ln2}−{I}_{{AB}} =\mathrm{ln2}−\left[−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\left(−\mathrm{ln2}\right)\right] \\$$ $${I}=\mathrm{ln2}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\mathrm{ln2}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\$$ $$\because{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({y}^{{y}} \right)^{\left({y}^{{y}} \right)^{\left({y}^{{y}} \right)} } {dy}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\$$ $${solution}\:{by}\:{mathdave} \\$$ $$\mathrm{15}/\mathrm{08}/\mathrm{2020} \\$$

Commented byHer_Majesty last updated on 18/Aug/20

$${I}\:{don}'{t}\:{think}\:{this}\:{is}\:{true}.\:{the}\:{value}\:{of}\:{the} \\$$ $${integral}\:{is}\:{slightly}\:{less}\:{than}\:\pi^{\mathrm{2}} /\mathrm{12} \\$$ $$\pi^{\mathrm{2}} /\mathrm{12}\approx.\mathrm{822467033424} \\$$ $${but}\:{I}\approx.\mathrm{820108872462} \\$$