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...](https://www.tinkutara.com/question/Q114721.png)
$$\:\:\:\:\:\:\:\:\:\:….{nice}\:\:{calculus}…. \\ $$$$\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Phi=\int_{\mathrm{0}\:} ^{\:\mathrm{1}} {xln}\left[{ln}\left({x}\right).{ln}\left(\mathrm{1}−{x}\right)\right]{dx}=−\gamma \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\gamma\:::=\:{euler}\:\:{mascheroni}\:{constant}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{m}.{n}.{july}.\mathrm{1970}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
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](https://www.tinkutara.com/question/Q114727.png)
$${by}\:{part} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right){ln}\left[{ln}\left(\mathrm{1}−{x}\right){ln}\left({x}\right)\right]{dx} \\ $$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {xln}\left({ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left[{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right).−{ln}\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$${in}\:\mathrm{2}{nd}\:\mathrm{1}−{x}={t}\Rightarrow \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx} \\ $$$${t}=−{ln}\left({x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){e}^{−{t}} {dt} \\ $$$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$\Gamma'\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {ln}\left({t}\right){e}^{−{t}} {dt}=\Psi\left(\mathrm{1}\right)=−\gamma \\ $$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {xln}\left[{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx}=−\mathrm{2}\gamma \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {xln}\left[{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)\right]{dx}=−\gamma \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 20/Sep/20
$$\:{excellent}\:{sir}..{thank}\:{you}\:… \\ $$
Commented by maths mind last updated on 20/Sep/20
$${withe}\:{pleasur} \\ $$
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](https://www.tinkutara.com/question/Q114736.png)
$${my}\:{solution}.. \\ $$$$\:\:\:\:\:\:\Phi\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left[\left(−{ln}\left({x}\right)\right)\left(−{ln}\:\left(\mathrm{1}−{x}\right)\right)\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left(−{ln}\left({x}\right)\right){dx}\:+\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left(−{ln}\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xln}\left(−{ln}\left({x}\right)\right){dx}\:+\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−{x}\right){ln}\left(−{ln}\left({x}\right)\right){dx}\: \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(−{ln}\left({x}\right)\right){dx}\overset{\left[−{ln}\left({x}\right)={t}\right]} {=}\:−\int_{\infty} ^{\:\:\mathrm{0}} {e}^{−{t}} {ln}\left({t}\right){dt} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{t}} {ln}\left({t}\right){dt}\:=\:−\gamma\:\:\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970} \\ $$$$\:\:\: \\ $$