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Question Number 125668 by mnjuly1970 last updated on 12/Dec/20
... advanced calculus... evaluate::: Φ=∫_0 ^( (π/4)) ln(ln(cot(x)))dx=?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:{evaluate}::: \\ $$$$\:\:\:\:\:\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left({ln}\left({cot}\left({x}\right)\right)\right){dx}=? \\ $$$$ \\ $$
Answered by mindispower last updated on 13/Dec/20
Φ=∫_1 ^∞ ((ln(ln(x)))/(1+x^2 ))dx=∫_0 ^∞ ((ln(t))/(1+e^(2t) ))e^t dt =∫_0 ^∞ ((ln(t)e^(−t) )/(1+e^(−2t) ))dt =Σ_(k≥0) ∫_0 ^∞ (−1)^k ln(t)e^(−(1+2k)t) ln(t)=∂_s t^s ∣s=0 g(s)=Σ_(k≥0) ∫_0 ^∞ (−1)^k t^s e^(−(1+2k)t) dt =Σ_(k≥0) ∫_0 ^∞ (−1)^k ((y^s e^(−y) dy)/((1+2k)^(s+1) )) =Σ_(k≥0) (((−1)^k )/((1+2k)^(s+1) ))∫_0 ^∞ y^s e^(−y) dy =Σ_(k≥0) (((−1)^k )/((1+2k)^(s+1) ))Γ(s+1) =Γ(s+1)Σ_(k≥0) ((1/((1+4k)^(s+1) ))−(1/((3+4k)^(s+1) ))) =((Γ(s+1))/4^(s+1) )(Σ_(k≥0) (1/((k+(1/4))^(s+1) ))−Σ_(k≥0) (1/((k+(3/4))^(s+1) ))) =((Γ(s+1))/4^(s+1) )(ζ(s+1,(1/4))−ζ(s+1,(3/4)) ζ(s,q)=Σ_(n≥0) (1/((n+q)^s )),Zets hrwitz function we want g′(0) ((Γ′(s+1)4^(s+1) −4^(s+1) ln(4)Γ(s+1))/4^(2s+2) )∣_(s=0) .lim_(s→0) (ζ(s+1,(1/4))−ζ(s+1,(3/4))) +lim_(s→0) ∂_s (ζ(s+1,(1/4))−ζ(s+1,(3/4))).((Γ(s+1))/4^(s+1) ) ζ(s+1,q)=(1/s)+Σ(((−1)^n )/(n!))γ_n (q)s^n ,laurent serie expansion γ_n (q) Stieltjes constante ζ(s+1,(1/4))−ζ(s+1,(3/4))=h(s)=Σ_(n≥0) (((−1)^n )/(n!))(γ_n ((1/4))−γ_n ((3/4)))s^n lim_(s→0) h(s)=γ_0 ((1/4))−γ_0 ((3/4)) ∂_s h(s)=Σ_(n≥1) (((−1)^n )/((n−1)!))(γ_n ((1/4))−γ_n ((3/4))s^(n−1) lim_(s→0) h′(s)=γ_1 ((3/4))−γ_1 ((1/4)) γ_1 ((3/4))−γ_1 (1−(3/4))=2πΣ_(l=1) ^3 sin(((6πl)/4))lnΓ((l/4))−π(γ+ln(8π))cot(((3π)/4)) =−2πΓ((1/4))+2πΓ((3/4))+π(γ+ln(8π)) =2π(Γ((3/4))−Γ((1/4)))+π(γ+ln(8π)) γ_0 ((1/4))−γ_0 ((3/4))=Ψ((3/4))−Ψ((1/4))=−πcot(((3π)/4)) =π g′(0)=((4Γ′(1)−4ln(4)Γ(1))/4^2 )π+(2π(Γ((3/4))−Γ((1/4)))+π(γ+ln(8π)).(1/4) =−(γ/4)π−((ln(2))/2)π+(π/2)(Γ((3/4))−Γ((1/4)))+((γπ)/4)+(π/4)ln(8π) =(π/4)ln(2π)+(π/2)(Γ((3/4))−Γ((1/4)))
$$\Phi=\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({ln}\left({x}\right)\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({t}\right)}{\mathrm{1}+{e}^{\mathrm{2}{t}} }{e}^{{t}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({t}\right){e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{2}{t}} }{dt} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} {ln}\left({t}\right){e}^{−\left(\mathrm{1}+\mathrm{2}{k}\right){t}} \\ $$$${ln}\left({t}\right)=\partial_{{s}} {t}^{{s}} \mid{s}=\mathrm{0} \\ $$$${g}\left({s}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} {t}^{{s}} {e}^{−\left(\mathrm{1}+\mathrm{2}{k}\right){t}} {dt} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \frac{{y}^{{s}} {e}^{−{y}} {dy}}{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{s}+\mathrm{1}} } \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{s}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {y}^{{s}} {e}^{−{y}} {dy} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{1}+\mathrm{2}{k}\right)^{{s}+\mathrm{1}} }\Gamma\left({s}+\mathrm{1}\right) \\ $$$$=\Gamma\left({s}+\mathrm{1}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{4}{k}\right)^{{s}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{3}+\mathrm{4}{k}\right)^{{s}+\mathrm{1}} }\right) \\ $$$$=\frac{\Gamma\left({s}+\mathrm{1}\right)}{\mathrm{4}^{{s}+\mathrm{1}} }\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{s}+\mathrm{1}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{{s}+\mathrm{1}} }\right) \\ $$$$=\frac{\Gamma\left({s}+\mathrm{1}\right)}{\mathrm{4}^{{s}+\mathrm{1}} }\left(\zeta\left({s}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{4}}\right)−\zeta\left({s}+\mathrm{1},\frac{\mathrm{3}}{\mathrm{4}}\right)\right. \\ $$$$\zeta\left({s},{q}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+{q}\right)^{{s}} },{Zets}\:{hrwitz}\:{function} \\ $$$${we}\:{want}\:{g}'\left(\mathrm{0}\right) \\ $$$$\frac{\Gamma'\left({s}+\mathrm{1}\right)\mathrm{4}^{{s}+\mathrm{1}} −\mathrm{4}^{{s}+\mathrm{1}} {ln}\left(\mathrm{4}\right)\Gamma\left({s}+\mathrm{1}\right)}{\mathrm{4}^{\mathrm{2}{s}+\mathrm{2}} }\mid_{{s}=\mathrm{0}} .\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\zeta\left({s}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{4}}\right)−\zeta\left({s}+\mathrm{1},\frac{\mathrm{3}}{\mathrm{4}}\right)\right) \\ $$$$+\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\partial_{{s}} \left(\zeta\left({s}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{4}}\right)−\zeta\left({s}+\mathrm{1},\frac{\mathrm{3}}{\mathrm{4}}\right)\right).\frac{\Gamma\left({s}+\mathrm{1}\right)}{\mathrm{4}^{{s}+\mathrm{1}} } \\ $$$$\zeta\left({s}+\mathrm{1},{q}\right)=\frac{\mathrm{1}}{{s}}+\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\gamma_{{n}} \left({q}\right){s}^{{n}} ,{laurent}\:{serie} \\ $$$${expansion}\:\gamma_{{n}} \left({q}\right)\:{Stieltjes}\:{constante} \\ $$$$\zeta\left({s}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{4}}\right)−\zeta\left({s}+\mathrm{1},\frac{\mathrm{3}}{\mathrm{4}}\right)={h}\left({s}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\left(\gamma_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\gamma_{{n}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right){s}^{{n}} \\ $$$$\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}{h}\left({s}\right)=\gamma_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\gamma_{\mathrm{0}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\partial_{{s}} {h}\left({s}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)!}\left(\gamma_{{n}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\gamma_{{n}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right){s}^{{n}−\mathrm{1}} \right. \\ $$$$\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}{h}'\left({s}\right)=\gamma_{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\gamma_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\gamma_{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\gamma_{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{2}\pi\underset{{l}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{sin}\left(\frac{\mathrm{6}\pi{l}}{\mathrm{4}}\right){ln}\Gamma\left(\frac{{l}}{\mathrm{4}}\right)−\pi\left(\gamma+{ln}\left(\mathrm{8}\pi\right)\right){cot}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right) \\ $$$$=−\mathrm{2}\pi\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{2}\pi\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)+\pi\left(\gamma+{ln}\left(\mathrm{8}\pi\right)\right) \\ $$$$=\mathrm{2}\pi\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)+\pi\left(\gamma+{ln}\left(\mathrm{8}\pi\right)\right) \\ $$$$\gamma_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\gamma_{\mathrm{0}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi{cot}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right) \\ $$$$=\pi \\ $$$${g}'\left(\mathrm{0}\right)=\frac{\mathrm{4}\Gamma'\left(\mathrm{1}\right)−\mathrm{4}{ln}\left(\mathrm{4}\right)\Gamma\left(\mathrm{1}\right)}{\mathrm{4}^{\mathrm{2}} }\pi+\left(\mathrm{2}\pi\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)+\pi\left(\gamma+{ln}\left(\mathrm{8}\pi\right)\right).\frac{\mathrm{1}}{\mathrm{4}}\right. \\ $$$$ \\ $$$$ \\ $$$$=−\frac{\gamma}{\mathrm{4}}\pi−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\pi+\frac{\pi}{\mathrm{2}}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right)+\frac{\gamma\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{8}\pi\right) \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\pi\right)+\frac{\pi}{\mathrm{2}}\left(\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$
Commented by mindispower last updated on 13/Dec/20
we can use Γ(z)Γ(z+(1/2))=2^(1−2z) (√π)Γ(2z) Γ((1/4)).Γ((3/4))=2^(1/2) (√π).Γ((1/2))=π(√2) ⇒Γ((3/4))=π(√2).(1/(Γ((1/4))))...to express using just Γ((1/4))
$${we}\:{can}\: \\ $$$${use}\:\Gamma\left({z}\right)\Gamma\left({z}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}^{\mathrm{1}−\mathrm{2}{z}} \sqrt{\pi}\Gamma\left(\mathrm{2}{z}\right) \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\pi}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\pi\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\pi\sqrt{\mathrm{2}}.\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}…{to}\:{express}\:{using} \\ $$$${just}\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 13/Dec/20
thanks alot mr power......
$${thanks}\:{alot}\: \\ $$$${mr}\:{power}…… \\ $$
Commented by mindispower last updated on 13/Dec/20
withe pleasur
$${withe}\:{pleasur}\: \\ $$

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