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Question Number 114721 by mnjuly1970 last updated on 20/Sep/20

          ....nice  calculus....    prove  that::                             Φ=∫_(0 ) ^( 1) xln[ln(x).ln(1−x)]dx=−γ                             γ ::= euler  mascheroni constant.                                         ...m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}.... \\ $$$$\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Phi=\int_{\mathrm{0}\:} ^{\:\mathrm{1}} {xln}\left[{ln}\left({x}\right).{ln}\left(\mathrm{1}−{x}\right)\right]{dx}=−\gamma \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:::=\:{euler}\:\:{mascheroni}\:{constant}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Answered by maths mind last updated on 20/Sep/20

by part  =∫_0 ^1 (1−x)ln[ln(1−x)ln(x)]dx  ⇔2∫_0 ^1 xln(ln(x)ln(1−x))dx=∫_0 ^1 ln[ln(x)ln(1−x)]dx  =∫_0 ^1 ln(−ln(x).−ln(1−x))dx  =∫_0 ^1 ln(−ln(x))dx+∫_0 ^1 ln(−ln(1−x))dx  in 2nd 1−x=t⇒  =2∫_0 ^1 ln(−ln(x))dx  t=−ln(x)  =2∫_0 ^∞ ln(t)e^(−t) dt  Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt  Γ′(1)=∫_0 ^∞ ln(t)e^(−t) dt=Ψ(1)=−γ  ⇔2∫_0 ^1 xln[ln(x)ln(1−x)]dx=2∫_0 ^1 ln(−ln(x))dx=−2γ  ⇒∫_0 ^1 xln[ln(x)ln(1−x)]dx=−γ

$${by}\:{part} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right){ln}\left[{ln}\left(\mathrm{1}−{x}\right){ln}\left({x}\right)\right]{dx} \\ $$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {xln}\left({ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left[{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right).−{ln}\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$${in}\:\mathrm{2}{nd}\:\mathrm{1}−{x}={t}\Rightarrow \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx} \\ $$$${t}=−{ln}\left({x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){e}^{−{t}} {dt} \\ $$$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\Gamma'\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){e}^{−{t}} {dt}=\Psi\left(\mathrm{1}\right)=−\gamma \\ $$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {xln}\left[{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx}=−\mathrm{2}\gamma \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {xln}\left[{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]{dx}=−\gamma \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 20/Sep/20

 excellent sir..thank you ...

$$\:{excellent}\:{sir}..{thank}\:{you}\:... \\ $$

Commented by maths mind last updated on 20/Sep/20

withe pleasur

$${withe}\:{pleasur} \\ $$

Answered by mnjuly1970 last updated on 20/Sep/20

my solution..        Φ =∫_0 ^( 1) xln[(−ln(x))(−ln (1−x))]dx  =∫_0 ^( 1) xln(−ln(x))dx +∫_0 ^( 1) xln(−ln(1−x))dx  =∫_0 ^( 1) xln(−ln(x))dx +∫_0 ^( 1) (1−x)ln(−ln(x))dx   =∫_0 ^( 1) ln(−ln(x))dx=^([−ln(x)=t])  −∫_∞ ^(  0) e^(−t) ln(t)dt   =∫_0 ^( ∞) e^(−t) ln(t)dt = −γ    ✓            m.n.july.1970

$${my}\:{solution}.. \\ $$$$\:\:\:\:\:\:\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left[\left(−{ln}\left({x}\right)\right)\left(−{ln}\:\left(\mathrm{1}−{x}\right)\right)\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left(−{ln}\left({x}\right)\right){dx}\:+\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left(−{ln}\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left(−{ln}\left({x}\right)\right){dx}\:+\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−{x}\right){ln}\left(−{ln}\left({x}\right)\right){dx}\: \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx}\overset{\left[−{ln}\left({x}\right)={t}\right]} {=}\:−\int_{\infty} ^{\:\:\mathrm{0}} {e}^{−{t}} {ln}\left({t}\right){dt} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {ln}\left({t}\right){dt}\:=\:−\gamma\:\:\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970} \\ $$$$\:\:\: \\ $$

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