Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 137501 by mnjuly1970 last updated on 03/Apr/21

....advanced .... calculus.... prove that:: 𝛗=∫_0 ^( 1) ((ln(x).ln(1βˆ’x))/(1+x))dx=((13)/8)ΞΆ(3)βˆ’(Ο€^2 /4)ln(2)....

$$\:\:\:\:\:\:\:....{advanced}\:....\:{calculus}.... \\ $$$$\:\:{prove}\:{that}:: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}\right).{ln}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}+{x}}{dx}=\frac{\mathrm{13}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\left(\mathrm{2}\right).... \\ $$

Answered by Ñï= last updated on 04/Apr/21

Ο†=∫_0 ^1 ((lnx ln(1βˆ’x))/(1+x))dx =∫_0 ^1 ((lnxln(1βˆ’x^2 ))/(1+x))βˆ’βˆ«_0 ^1 ((lnxln(1+x))/(1+x))dx =∫_0 ^1 ((lnxln(1βˆ’x^2 ))/(1βˆ’x^2 ))dxβˆ’βˆ«_0 ^1 ((xln xln (1βˆ’x^2 ))/(1βˆ’x^2 ))dxβˆ’βˆ«_0 ^1 ((ln xln (1+x))/(1+x))dx =(βˆ‚/(βˆ‚aβˆ‚b))∣_(a=0,b=βˆ’1) ∫_0 ^1 x^a (1βˆ’x^2 )^b dxβˆ’(βˆ‚/(βˆ‚aβˆ‚b))∣_(a=1,b=βˆ’1) ∫_0 ^1 x^a (1βˆ’x^2 )^b dxβˆ’{∫((ln(1+x){ln(1+x)+ln(x/(1+x))})/(1+x))}_0 ^1 =(βˆ‚/(βˆ‚aβˆ‚b))∣_(a=0,b=βˆ’1) (1/2)∫_0 ^1 u^((aβˆ’1)/2) (1βˆ’u)^b duβˆ’(βˆ‚/(βˆ‚aβˆ‚b))∣_(a=1,b=βˆ’1) (1/2)∫_0 ^1 u^((aβˆ’1)/2) (1βˆ’u)^b duβˆ’{∫((ln^2 (1+x))/(1+x))dx+∫((ln(1βˆ’(1/(1+x))))/(1+x))ln (1+x)dx}_0 ^1 =(βˆ‚/(βˆ‚aβˆ‚b))∣_(a=0,b=βˆ’1) (1/2)B(((a+1)/2),b+1)βˆ’(βˆ‚/(βˆ‚aβˆ‚b))∣_(a=1,b=βˆ’1) (1/2)B(((a+1)/2),b+1)βˆ’(1/3)ln^3 2+Ξ£_(n=1) ^∞ ∫_0 ^1 (((1+x)^(βˆ’n) )/n)βˆ™((ln (1+x))/(1+x))dx =(3/2)ΞΆ(3)βˆ’(Ο€^2 /4)ln2βˆ’(1/3)ln^3 2+Ξ£_(n=1) ^∞ (1/n){∫_0 ^1 (1+x)^(βˆ’nβˆ’1) ln(1+x)dx} =(3/2)ΞΆ(3)βˆ’(Ο€^2 /4)ln2βˆ’(1/3)ln^3 2+Ξ£_(n=1) ^∞ (1/n){(((1+x)^(βˆ’n) )/(βˆ’n))ln(1+x)βˆ’βˆ«(((1+x)^(βˆ’nβˆ’1) )/(βˆ’n))dx}_0 ^1 =(3/2)ΞΆ(3)βˆ’(Ο€^2 /4)ln2βˆ’(1/3)ln^3 2βˆ’Ξ£_(n=1) ^∞ (((1+x)^(βˆ’n) ln(1+x))/n^2 )∣_0 ^1 +Ξ£_(n=1) ^∞ (((1+x)^(βˆ’n) )/n^3 )∣_0 ^1 =(3/2)ΞΆ(3)βˆ’(Ο€^2 /4)ln2βˆ’(1/3)ln^3 2βˆ’Li_2 ((1/(1+x)))ln(1+x)∣_0 ^1 βˆ’Li_3 ((1/(1+x)))∣_0 ^1 =(3/2)ΞΆ(3)βˆ’(Ο€^2 /4)ln2βˆ’(1/3)ln^3 2βˆ’ln2Li_2 ((1/2))βˆ’Li_3 ((1/2))+Li_3 (1) =ans...

$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}\:{ln}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnxln}\left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }{dx}βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\:{xln}\:\left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }{dx}βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\:{xln}\:\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{0},{b}=βˆ’\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)^{{b}} {dx}βˆ’\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{1},{b}=βˆ’\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)^{{b}} {dx}βˆ’\left\{\int\frac{{ln}\left(\mathrm{1}+{x}\right)\left\{{ln}\left(\mathrm{1}+{x}\right)+{ln}\frac{{x}}{\mathrm{1}+{x}}\right\}}{\mathrm{1}+{x}}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{0},{b}=βˆ’\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\left({a}βˆ’\mathrm{1}\right)/\mathrm{2}} \left(\mathrm{1}βˆ’{u}\right)^{{b}} {du}βˆ’\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{1},{b}=βˆ’\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\left({a}βˆ’\mathrm{1}\right)/\mathrm{2}} \left(\mathrm{1}βˆ’{u}\right)^{{b}} {du}βˆ’\left\{\int\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx}+\int\frac{{ln}\left(\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)}{\mathrm{1}+{x}}{ln}\:\left(\mathrm{1}+{x}\right){dx}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{0},{b}=βˆ’\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}},{b}+\mathrm{1}\right)βˆ’\frac{\partial}{\partial{a}\partial{b}}\mid_{{a}=\mathrm{1},{b}=βˆ’\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}},{b}+\mathrm{1}\right)βˆ’\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}+{x}\right)^{βˆ’{n}} }{{n}}\centerdot\frac{{ln}\:\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}βˆ’\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left\{\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{x}\right)^{βˆ’{n}βˆ’\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right){dx}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}βˆ’\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\left\{\frac{\left(\mathrm{1}+{x}\right)^{βˆ’{n}} }{βˆ’{n}}{ln}\left(\mathrm{1}+{x}\right)βˆ’\int\frac{\left(\mathrm{1}+{x}\right)^{βˆ’{n}βˆ’\mathrm{1}} }{βˆ’{n}}{dx}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}βˆ’\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}+{x}\right)^{βˆ’{n}} {ln}\left(\mathrm{1}+{x}\right)}{{n}^{\mathrm{2}} }\mid_{\mathrm{0}} ^{\mathrm{1}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}+{x}\right)^{βˆ’{n}} }{{n}^{\mathrm{3}} }\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}βˆ’\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}βˆ’{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){ln}\left(\mathrm{1}+{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} βˆ’{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\zeta\left(\mathrm{3}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{ln}\mathrm{2}βˆ’\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \mathrm{2}βˆ’{ln}\mathrm{2}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)βˆ’{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{Li}_{\mathrm{3}} \left(\mathrm{1}\right) \\ $$$$={ans}... \\ $$

Commented by mnjuly1970 last updated on 04/Apr/21

great ...thanks alot...

$${great}\:...{thanks}\:{alot}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com