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Question Number 122330 by mnjuly1970 last updated on 15/Nov/20

...advanced calculus... prove that : Re(∫_0 ^( (π/2)) sin^3 (x)ln(ln(cos(x)))dx) =^? ((ln(3)−2γ)/3) ✓

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{advanced}\:\:{calculus}... \\ $$$$\:\:{prove}\:{that}\:: \\ $$$$\:\:\:\mathscr{R}{e}\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{3}} \left({x}\right){ln}\left({ln}\left({cos}\left({x}\right)\right)\right){dx}\right) \\ $$$$\:\:\:\:\:\:\:\overset{?} {=}\frac{{ln}\left(\mathrm{3}\right)−\mathrm{2}\gamma}{\mathrm{3}}\:\checkmark \\ $$

Answered by mathmax by abdo last updated on 16/Nov/20

from where you get this sir mn...

$$\mathrm{from}\:\mathrm{where}\:\mathrm{you}\:\mathrm{get}\:\mathrm{this}\:\mathrm{sir}\:\mathrm{mn}... \\ $$

Commented by mathmax by abdo last updated on 16/Nov/20

thank you sir

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

Commented by mnjuly1970 last updated on 16/Nov/20

hi mr max from App pimterest and brilliant math in google...

$${hi}\:{mr}\:{max} \\ $$$${from} \\ $$$${App}\:{pimterest}\: \\ $$$$\:{and}\: \\ $$$${brilliant}\:\:{math}\:{in}\:{google}... \\ $$

Answered by mindispower last updated on 16/Nov/20

let cos(x)=t⇒sin(x)dx=−dt ∫_0 ^(π/2) sin^2 (x)ln(ln(cos(x)))sin(x)dx =∫_0 ^1 (1−t^2 )ln(ln(t))dt −ln(t)=u⇒dt=−e^(−u) du Ω=∫_0 ^∞ (1−e^(−2u) )ln(−u)e^(−u) du ln(−u)=ln(u)+iπ Re(Ω)=∫_0 ^∞ e^(−u) ln(u)du−∫_0 ^∞ e^(−3u) ln(u)du...I 3u=y⇒du=(dy/3) ⇔I=(2/3)∫_0 ^∞ e^(−u) ln(u)du+(1/3)∫_0 ^∞ e^(−y) ln(3)dy Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt⇒Γ(1)=∫_0 ^∞ e^(−t) dt Γ′(x)=∫_0 ^∞ ln(t)t^(x−1) ln(t)dt Γ′(1)=∫_0 ^∞ ln(t)e^(−t) dt Re(Ω)=(2/3)Γ′(1)+(1/3)Γ(1) Γ′(1)=Ψ(1)=−γ ⇔Re(Ω)=−(2/3)γ+(1/3)=((1−2γ)/3)

$${let}\:{cos}\left({x}\right)={t}\Rightarrow{sin}\left({x}\right){dx}=−{dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left({x}\right){ln}\left({ln}\left({cos}\left({x}\right)\right)\right){sin}\left({x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){ln}\left({ln}\left({t}\right)\right){dt} \\ $$$$−{ln}\left({t}\right)={u}\Rightarrow{dt}=−{e}^{−{u}} {du} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−{e}^{−\mathrm{2}{u}} \right){ln}\left(−{u}\right){e}^{−{u}} {du} \\ $$$${ln}\left(−{u}\right)={ln}\left({u}\right)+{i}\pi \\ $$$${Re}\left(\Omega\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{u}} {ln}\left({u}\right){du}−\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{3}{u}} {ln}\left({u}\right){du}...{I} \\ $$$$\mathrm{3}{u}={y}\Rightarrow{du}=\frac{{dy}}{\mathrm{3}} \\ $$$$\Leftrightarrow{I}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}} {ln}\left({u}\right){du}+\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {ln}\left(\mathrm{3}\right){dy} \\ $$$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}\Rightarrow\Gamma\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {dt} \\ $$$$\Gamma'\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){t}^{{x}−\mathrm{1}} {ln}\left({t}\right){dt} \\ $$$$\Gamma'\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){e}^{−{t}} {dt} \\ $$$${Re}\left(\Omega\right)=\frac{\mathrm{2}}{\mathrm{3}}\Gamma'\left(\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{3}}\Gamma\left(\mathrm{1}\right) \\ $$$$\Gamma'\left(\mathrm{1}\right)=\Psi\left(\mathrm{1}\right)=−\gamma \\ $$$$\Leftrightarrow{Re}\left(\Omega\right)=−\frac{\mathrm{2}}{\mathrm{3}}\gamma+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}−\mathrm{2}\gamma}{\mathrm{3}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 16/Nov/20

thank you so much sir max

$${thank}\:{you}\:{so}\:{much}\:{sir}\:{max}\: \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 16/Nov/20

grateful sir midspower.very nice solution....

$${grateful}\:{sir}\:{midspower}.{very}\:{nice} \\ $$$${solution}.... \\ $$

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