Question-133452

Question Number 133452 by bagjagunawan last updated on 22/Feb/21

Answered by mnjuly1970 last updated on 22/Feb/21

𝛗=^(x=2y−1) ∫_((1/2) ) ^( 1) ((ln(2−2y)ln(ln(2y)))/(2y))(2)dy =∫_(1/2) ^( 1) (ln(2)+ln(1−y))(ln(2)+ln(y))(dy/y) =∫_(1/2) ^( 1) ((ln^2 (2)+ln(y).ln(2)+ln(2).ln(1−y)+ln(1−y).ln(y))/y) dy =[ln(y)ln(2)]_(1/2) ^1 +(1/2)[ln(2).ln^2 y]_(1/2) ^1 +ln(2)[li_((2)) ((1/2))−li_2 (1)]+∫_(1/2) ^( 1) ((ln(y)ln(1−y))/y)dy =ln^2 (2)−(1/2)ln^3 (2)+ln(2)[(π^2 /(12))−(1/2)ln^2 (2)−(π^2 /6)]−Φ....(∗) Φ=[−li_2 (y).ln(y)]_(1/2) ^1 +∫_((1/2) ) ^( 1) ((li_2 (y))/y)dy =−li_2 ((1/2))+li_3 (1)−li_3 ((1/2)) =((−π^2 )/(12))+(1/2)ln^2 (2)+ζ(3)−[(7/8)ζ(3)+(1/6)ln^3 (2)−(π^2 /(12))ln(2)] =(1/8)ζ(3)+(1/2)ln^2 (2)+(1/6)ln^3 (2)−(π^2 /(12)) +(π^2 /(12))ln(2) (∗∗) replacing (∗) → (∗∗)

$$\boldsymbol{\phi}\overset{{x}=\mathrm{2}{y}−\mathrm{1}} {=}\int_{\frac{\mathrm{1}}{\mathrm{2}}\:} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{2}−\mathrm{2}{y}\right){ln}\left({ln}\left(\mathrm{2}{y}\right)\right)}{\mathrm{2}{y}}\left(\mathrm{2}\right){dy} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \left({ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−{y}\right)\right)\left({ln}\left(\mathrm{2}\right)+{ln}\left({y}\right)\right)\frac{{dy}}{{y}} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{ln}\left({y}\right).{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}\right).{ln}\left(\mathrm{1}−{y}\right)+{ln}\left(\mathrm{1}−{y}\right).{ln}\left({y}\right)}{{y}}\:{dy}\:\:\: \\ $$$$=\left[{ln}\left({y}\right){ln}\left(\mathrm{2}\right)\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{2}\right).{ln}^{\mathrm{2}} {y}\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} +{ln}\left(\mathrm{2}\right)\left[{li}_{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{li}_{\mathrm{2}} \left(\mathrm{1}\right)\right]+\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\:\mathrm{1}} \frac{{ln}\left({y}\right){ln}\left(\mathrm{1}−{y}\right)}{{y}}{dy} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)+{ln}\left(\mathrm{2}\right)\left[\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right]−\Phi….\left(\ast\right) \\ $$$$\Phi=\left[−{li}_{\mathrm{2}} \left({y}\right).{ln}\left({y}\right)\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:+\int_{\frac{\mathrm{1}}{\mathrm{2}}\:} ^{\:\mathrm{1}} \frac{{li}_{\mathrm{2}} \left({y}\right)}{{y}}{dy} \\ $$$$=−{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{li}_{\mathrm{3}} \left(\mathrm{1}\right)−{li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\zeta\left(\mathrm{3}\right)−\left[\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}{ln}\left(\mathrm{2}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}{ln}\left(\mathrm{2}\right)\:\:\left(\ast\ast\right) \\ $$$$\:{replacing}\:\:\:\left(\ast\right)\:\rightarrow\:\left(\ast\ast\right) \\ $$$$ \\ $$

Commented by bagjagunawan last updated on 22/Feb/21

Thank you Sir what the “li” and ζ? I don′t know the syimbol

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$$${what}\:{the}\:“{li}''\:\mathrm{and}\:\zeta? \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{syimbol} \\ $$

Commented by mnjuly1970 last updated on 22/Feb/21

you are welcome li_2 (x)=_(function) ^(dilogarithm) Σ_(n=1) ^∞ (x^n /n^2 )=−∫_0 ^( x) ((ln(1−t))/t)dt similarly li_3 (x)=Σ_(n=1) ^∞ (x^n /n^3 )=−∫_(0 ) ^( x) ((ln(1−u))/u) du immidiate results li_3 (1)=Σ_(n=1 ) ^∞ (1/n^3 )=ζ(3)(Apery′s constant) li_2 (1)=Σ_(n=1) ^∞ (1/n^2 )=ζ(2)=(π^2 /6) li_4 (1)=Σ_(n=1) ^∞ (1/n^4 )=(π^4 /(90)) li_2 ((1/2))=(π^2 /(12))−(1/2)ln^2 (2) ,..... ζ (s)=Σ_(n=1) ^∞ (1/n^s ) (reiman−zeta function)..

$${you}\:{are}\:{welcome} \\ $$$$\:\:{li}_{\mathrm{2}} \left({x}\right)\underset{{function}} {\overset{{dilogarithm}} {=}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }=−\int_{\mathrm{0}} ^{\:{x}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$$\:{similarly}\:\:{li}_{\mathrm{3}} \left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{3}} }=−\int_{\mathrm{0}\:} ^{\:{x}} \frac{{ln}\left(\mathrm{1}−{u}\right)}{{u}}\:{du} \\ $$$${immidiate}\:\:{results} \\ $$$$\:\:{li}_{\mathrm{3}} \left(\mathrm{1}\right)=\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right)\left({Apery}'{s}\:{constant}\right) \\ $$$$\:{li}_{\mathrm{2}} \left(\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\:\:{li}_{\mathrm{4}} \left(\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\: \\ $$$$\:\:{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:,….. \\ $$$$\:\:\:\zeta\:\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{s}} }\:\:\left({reiman}−{zeta}\:{function}\right).. \\ $$

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