Question Number 123547 by mnjuly1970 last updated on 26/Nov/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:\:\:\:{calculus}.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\left\{\frac{{log}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}{log}^{\mathrm{3}} \left(\mathrm{2}\right)\checkmark \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 26/Nov/20
) −(1/2)log^2 (2) ✓ ii:: li_3 ((1/2))[ =_(prove it.) ^(why??) ](7/8)ζ(3)+(1/6)log^3 (2)−(π^2 /(12))log(2) ✓ iii:: li_3 (1) =ζ(3)✓ I=^(1−x=t) ∫_(1/2) ^( 1) ((log^2 (t))/(1−t))dt=^(i.b.p) [−log(1−t)log^2 (t)]_(1/2) ^1 +2∫_(1/2) ^( 1) ((log(t)log(1−t))/t)dt =log((1/2))log^2 ((1/2))+2∫_(1/2) ^( 1) log(t)d(−li_2 (t)) =−log^3 (2)+2{[log(t)li_2 (t)]_((1/2) ) ^1 +∫_(1/2) ^( 1) ((li_2 (t))/t)dt} =−log^3 (2)+2[log((1/2))li_2 ((1/2))]+2li_3 (1)−2li_3 ((1/2)) =−log^3 (2)−2log(2)((π^2 /(12))−(1/2)log^2 (2))+_↫ −^↬ (7/4)ζ(3)−(1/3)log^3 (2)+(π^2 /6)log(2)+2ζ(3) =((ζ (3 ))/4) −(1/3)log^3 (2) ✓ ✓ nice.calculus. m.n.july.1970](Q123590.png)
$$\:\:\:\:\:\:{solution}: \\ $$$$\:\:\:\:\:\:{special}\:{values}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{i}::\:{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\:\underset{{try}\:{it}.} {\overset{{why}??} {=}}\:\right]\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:−\frac{\mathrm{1}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{2}\right)\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left[\:\underset{{prove}\:{it}.} {\overset{{why}??} {=}}\:\right]\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{6}}{log}^{\mathrm{3}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}{log}\left(\mathrm{2}\right)\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{iii}::\:{li}_{\mathrm{3}} \left(\mathrm{1}\right)\:=\zeta\left(\mathrm{3}\right)\checkmark \\ $$$$\:\:\:\:\:\:\:\:\mathrm{I}\overset{\mathrm{1}−{x}={t}} {=}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \frac{{log}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}−{t}}{dt}\overset{{i}.{b}.{p}} {=}\left[−{log}\left(\mathrm{1}−{t}\right){log}^{\mathrm{2}} \left({t}\right)\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} +\mathrm{2}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \frac{{log}\left({t}\right){log}\left(\mathrm{1}−{t}\right)}{{t}}{dt}\: \\ $$$$\:\:\:\:\:\:={log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){log}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} {log}\left({t}\right){d}\left(−{li}_{\mathrm{2}} \left({t}\right)\right) \\ $$$$\:\:\:\:\:\:\:=−{log}^{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{2}\left\{\left[{log}\left({t}\right){li}_{\mathrm{2}} \left({t}\right)\right]_{\frac{\mathrm{1}}{\mathrm{2}}\:} ^{\mathrm{1}} +\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt}\right\} \\ $$$$\:\:\:\:\:\:\:=−{log}^{\mathrm{3}} \left(\mathrm{2}\right)+\mathrm{2}\left[{log}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right]+\mathrm{2}{li}_{\mathrm{3}} \left(\mathrm{1}\right)−\mathrm{2}{li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:=−{log}^{\mathrm{3}} \left(\mathrm{2}\right)−\mathrm{2}{log}\left(\mathrm{2}\right)\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{2}\right)\right)\underset{\looparrowleft} {+} \\ $$$$\:\:\:\:\:\:\:\:\:\overset{\looparrowright} {−}\:\frac{\mathrm{7}}{\mathrm{4}}\zeta\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{3}}{log}^{\mathrm{3}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{log}\left(\mathrm{2}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\zeta\:\left(\mathrm{3}\:\right)}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{3}}{log}^{\mathrm{3}} \left(\mathrm{2}\right)\:\:\checkmark\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{nice}.{calculus}.\:{m}.{n}.{july}.\mathrm{1970} \\ $$
Answered by mathmax by abdo last updated on 26/Nov/20
![I =∫_0 ^(1/2) ((log^2 (1−x))/x)dx ⇒I =_(x=(t/2)) 2∫_0 ^1 ((log^2 (1−(t/2)))/t)(dt/2) =∫_0 ^1 ((log^2 (1−(t/2)))/t) dt let f(a) =∫_0 ^1 ((log^2 (1−at))/t)dt[with 0<a<1 f^′ (a) =2∫_0 ^1 ((−t)/(1−at))((log(1−at))/t)dt =−2 ∫_0 ^1 ((log(1−at))/(1−at))dt =−2 ∫_0 ^1 log(1−at)Σ_(n=0) ^∞ a^n t^n dt =−2 Σ_(n=0) ^∞ a^n ∫_0 ^1 t^n log(1−at)dt =−2Σ_(n=0) ^∞ a^n U_n U_n =∫_0 ^1 t^n log(1−at)dt =[(t^(n+1) /(n+1))log(1−at)]_0 ^1 =((log(1−a))/(n+1)) −∫_0 ^1 (t^(n+1) /(n+1))×((−a)/(1−at))dt =((log(1−a))/(n+1)) +(a/(n+1))∫_0 ^1 (t^(n+1) /(1−at))dt and ∫_0 ^1 (t^(n+1) /(1−at))dt =_(1−at=z) ∫_1 ^(1−a) (((((1−z)/a))^(n+1) )/z)(−(1/a))dz =(1/a^(n+2) )∫_(1−a) ^a (((1−z)^(n+1) )/z)dz =(1/a^(n+2) )∫_(1−a) ^a (1/z)Σ_(k=0) ^(n+1) C_(n+1) ^k (−z)^k dz =(1/a^(n+2) )∫_(1−a) ^a Σ_(k=0) ^(n+1) C_(n+1) ^k (−1)^k z^(k−1) dz =(1/a^(n+2) )Σ_(k=0) ^(n+1) C_(n+1) ^k (−1)^k [(z^k /k)]_(1−a) ^a =(1/a^(n+2) )Σ_(k=0) ^(n+1) C_(n+1) ^k (((−1)^k )/k){a^k −(1−a)^k }....be continued...](Q123628.png)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=_{\mathrm{x}=\frac{\mathrm{t}}{\mathrm{2}}} \:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{t}}\:\mathrm{dt}\:\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{at}\right)}{\mathrm{t}}\mathrm{dt}\left[\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1}\right. \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{t}}{\mathrm{1}−\mathrm{at}}\frac{\mathrm{log}\left(\mathrm{1}−\mathrm{at}\right)}{\mathrm{t}}\mathrm{dt}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}−\mathrm{at}\right)}{\mathrm{1}−\mathrm{at}}\mathrm{dt} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}−\mathrm{at}\right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{a}^{\mathrm{n}} \:\mathrm{t}^{\mathrm{n}} \:\mathrm{dt} \\ $$$$=−\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{a}^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}} \mathrm{log}\left(\mathrm{1}−\mathrm{at}\right)\mathrm{dt}\:=−\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{a}^{\mathrm{n}} \:\mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}} \:\mathrm{log}\left(\mathrm{1}−\mathrm{at}\right)\mathrm{dt}\:=\left[\frac{\mathrm{t}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\mathrm{log}\left(\mathrm{1}−\mathrm{at}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{log}\left(\mathrm{1}−\mathrm{a}\right)}{\mathrm{n}+\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}×\frac{−\mathrm{a}}{\mathrm{1}−\mathrm{at}}\mathrm{dt}\:=\frac{\mathrm{log}\left(\mathrm{1}−\mathrm{a}\right)}{\mathrm{n}+\mathrm{1}}\:+\frac{\mathrm{a}}{\mathrm{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{at}}\mathrm{dt}\:\mathrm{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{at}}\mathrm{dt}\:\:=_{\mathrm{1}−\mathrm{at}=\mathrm{z}} \:\:\int_{\mathrm{1}} ^{\mathrm{1}−\mathrm{a}} \:\frac{\left(\frac{\mathrm{1}−\mathrm{z}}{\mathrm{a}}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{z}}\left(−\frac{\mathrm{1}}{\mathrm{a}}\right)\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}+\mathrm{2}} }\int_{\mathrm{1}−\mathrm{a}} ^{\mathrm{a}} \:\:\frac{\left(\mathrm{1}−\mathrm{z}\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{z}}\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}+\mathrm{2}} }\int_{\mathrm{1}−\mathrm{a}} ^{\mathrm{a}} \frac{\mathrm{1}}{\mathrm{z}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}+\mathrm{1}} \mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{k}} \left(−\mathrm{z}\right)^{\mathrm{k}} \mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}+\mathrm{2}} }\int_{\mathrm{1}−\mathrm{a}} ^{\mathrm{a}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}+\mathrm{1}} \mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{z}^{\mathrm{k}−\mathrm{1}} \:\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}+\mathrm{2}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}+\mathrm{1}} \:\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \:\left[\frac{\mathrm{z}^{\mathrm{k}} }{\mathrm{k}}\right]_{\mathrm{1}−\mathrm{a}} ^{\mathrm{a}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{n}+\mathrm{2}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}+\mathrm{1}} \:\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{k}} \frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\left\{\mathrm{a}^{\mathrm{k}} −\left(\mathrm{1}−\mathrm{a}\right)^{\mathrm{k}} \right\}....\mathrm{be}\:\mathrm{continued}... \\ $$