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Question Number 135821 by mnjuly1970 last updated on 16/Mar/21
             ....nice   .....   calculus....       prove that ::      𝛗=∫_0 ^( 1) (((ln(1βˆ’x))/(1βˆ’(√(1βˆ’x)))))dx=4(1βˆ’ΞΆ(2))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:\:…..\:\:\:{calculus}….\: \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}βˆ’\sqrt{\mathrm{1}βˆ’{x}}}\right){dx}=\mathrm{4}\left(\mathrm{1}βˆ’\zeta\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 16/Mar/21
Ξ¦=∫_0 ^1  ((log(1βˆ’x))/(1βˆ’(√(1βˆ’x))))dx  we do the changement(√(1βˆ’x))=t β‡’1βˆ’x=t^2  β‡’x=1βˆ’t^2   Ξ¦ =βˆ’βˆ«_0 ^1 ((log(1βˆ’1+t^2 ))/(1βˆ’t))(βˆ’2t)dt =4∫_0 ^1  ((tlog(t))/(1βˆ’t))dt  =4∫_0 ^1  tlogtΞ£_(n=0) ^∞  t^n  dt =4Ξ£_(n=0) ^∞  ∫_0 ^1  t^(n+1) logt dt =4Ξ£ U_n   U_n =∫_0 ^1  t^(n+1) logt dt =[(t^(n+2) /(n+2))logt]_0 ^1 βˆ’(1/(n+2))∫_0 ^1  t^(n+1)  dt  =βˆ’(1/((n+2)^2 )) β‡’Ξ¦ =βˆ’4 Ξ£_(n=0) ^∞  (1/((n+2)^2 ))  =βˆ’4Ξ£_(n=2) ^∞  (1/n^2 ) =βˆ’4{Ξ£_(n=1) ^∞  (1/n^2 )βˆ’1}  =4βˆ’4.(Ο€^2 /6) =4βˆ’((2Ο€^2 )/3)
$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}βˆ’\mathrm{x}\right)}{\mathrm{1}βˆ’\sqrt{\mathrm{1}βˆ’\mathrm{x}}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\sqrt{\mathrm{1}βˆ’\mathrm{x}}=\mathrm{t}\:\Rightarrow\mathrm{1}βˆ’\mathrm{x}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\mathrm{1}βˆ’\mathrm{t}^{\mathrm{2}} \\ $$$$\Phi\:=βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{log}\left(\mathrm{1}βˆ’\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}βˆ’\mathrm{t}}\left(βˆ’\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{tlog}\left(\mathrm{t}\right)}{\mathrm{1}βˆ’\mathrm{t}}\mathrm{dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{tlogt}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{n}} \:\mathrm{dt}\:=\mathrm{4}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}+\mathrm{1}} \mathrm{logt}\:\mathrm{dt}\:=\mathrm{4}\Sigma\:\mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}+\mathrm{1}} \mathrm{logt}\:\mathrm{dt}\:=\left[\frac{\mathrm{t}^{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\mathrm{logt}\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}+\mathrm{1}} \:\mathrm{dt} \\ $$$$=βˆ’\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow\Phi\:=βˆ’\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=βˆ’\mathrm{4}\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:=βˆ’\mathrm{4}\left\{\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }βˆ’\mathrm{1}\right\} \\ $$$$=\mathrm{4}βˆ’\mathrm{4}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\mathrm{4}βˆ’\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$
Answered by Dwaipayan Shikari last updated on 16/Mar/21
∫_0 ^1 ((log(1βˆ’x))/(1βˆ’(√(1βˆ’x))))dx=∫_0 ^1 ((log(x))/(1βˆ’(√x)))dx  =∫_0 ^1 ((4tlog(t))/(1βˆ’t))dt  =4Ξ¦β€²(2)=4(1βˆ’(Ο€^2 /6))  Ξ¦(Ξ±)=∫_0 ^1 ((1βˆ’x^(Ξ±βˆ’1) )/(1βˆ’x))dx=βˆ’Ξ³+ψ(Ξ±)  Ξ¦β€²(Ξ±)=βˆ’βˆ«_0 ^1 ((x^(Ξ±βˆ’1) log(x))/(1βˆ’x))dx=ψ^1 (Ξ±)β‡’βˆ«_0 ^1 ((xlog(x))/(1βˆ’x))=βˆ’Οˆ^1 (2)  Ξ¦β€²(2)=βˆ’Ξ£_(n=1) ^∞ (1/((n+1)^2 ))=βˆ’((Ο€^2 /6)βˆ’1)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}βˆ’\sqrt{\mathrm{1}βˆ’{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({x}\right)}{\mathrm{1}βˆ’\sqrt{{x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}{tlog}\left({t}\right)}{\mathrm{1}βˆ’{t}}{dt} \\ $$$$=\mathrm{4}\Phi'\left(\mathrm{2}\right)=\mathrm{4}\left(\mathrm{1}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$\Phi\left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}βˆ’{x}^{\alphaβˆ’\mathrm{1}} }{\mathrm{1}βˆ’{x}}{dx}=βˆ’\gamma+\psi\left(\alpha\right) \\ $$$$\Phi'\left(\alpha\right)=βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alphaβˆ’\mathrm{1}} {log}\left({x}\right)}{\mathrm{1}βˆ’{x}}{dx}=\psi^{\mathrm{1}} \left(\alpha\right)\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xlog}\left({x}\right)}{\mathrm{1}βˆ’{x}}=βˆ’\psi^{\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\Phi'\left(\mathrm{2}\right)=βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=βˆ’\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}βˆ’\mathrm{1}\right) \\ $$
Commented by mnjuly1970 last updated on 16/Mar/21
thanks alot mr payan...
$${thanks}\:{alot}\:{mr}\:{payan}… \\ $$
Answered by Ñï= last updated on 16/Mar/21
∫_0 ^1 ((ln(1βˆ’x))/(1βˆ’(√(1βˆ’x))))dx=∫_0 ^1 ((lnx)/(1βˆ’(√x)))dx=∫_0 ^1 (((1βˆ’(√x))lnx)/(1βˆ’x))dx  =∫_0 ^1 ((lnx)/(1βˆ’x))dxβˆ’βˆ«_0 ^1 (((√x)lnx)/(1βˆ’x))dx  =∫_0 ^1 ((lnx)/(1βˆ’x))dxβˆ’βˆ«_0 ^1 ((4x^2 lnx)/(1βˆ’x^2 ))dx      ((√x)β†’x)  =βˆ’Li_2 (1)βˆ’4∫_0 ^1 (1βˆ’(1/(1βˆ’x^2 )))lnxdx  =βˆ’Li_2 (1)βˆ’4∫_0 ^1 lnxdx+4∫_0 ^1 ((lnx)/((1βˆ’x)(1+x)))dx  =βˆ’Li_2 (1)+4+2∫_0 ^1 ((1/(1βˆ’x))+(1/(1+x)))lnxdx  =βˆ’Li_2 (1)+4βˆ’2Li_2 (1)+2Li_2 (βˆ’1)  =βˆ’4Li_2 (1)+4  =4(1βˆ’ΞΆ(2))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}βˆ’{x}\right)}{\mathrm{1}βˆ’\sqrt{\mathrm{1}βˆ’{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}βˆ’\sqrt{{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}βˆ’\sqrt{{x}}\right){lnx}}{\mathrm{1}βˆ’{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}βˆ’{x}}{dx}βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{{x}}{lnx}}{\mathrm{1}βˆ’{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}βˆ’{x}}{dx}βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}{x}^{\mathrm{2}} {lnx}}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\left(\sqrt{{x}}\rightarrow{x}\right) \\ $$$$=βˆ’{Li}_{\mathrm{2}} \left(\mathrm{1}\right)βˆ’\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{1}βˆ’{x}^{\mathrm{2}} }\right){lnxdx} \\ $$$$=βˆ’{Li}_{\mathrm{2}} \left(\mathrm{1}\right)βˆ’\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {lnxdx}+\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\left(\mathrm{1}βˆ’{x}\right)\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=βˆ’{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{4}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}βˆ’{x}}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){lnxdx} \\ $$$$=βˆ’{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{4}βˆ’\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{2}{Li}_{\mathrm{2}} \left(βˆ’\mathrm{1}\right) \\ $$$$=βˆ’\mathrm{4}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{4} \\ $$$$=\mathrm{4}\left(\mathrm{1}βˆ’\zeta\left(\mathrm{2}\right)\right) \\ $$
Commented by mnjuly1970 last updated on 16/Mar/21
  thank you do much...
$$\:\:{thank}\:{you}\:{do}\:{much}… \\ $$
Answered by mnjuly1970 last updated on 16/Mar/21
     solution:    y^2 =1βˆ’x    𝛗=2∫_0 ^( 1) ((ln((√(1βˆ’x)) ))/(1βˆ’(√(1βˆ’x))))dx      𝛗=4∫_0 ^( 1) ((yln(y))/(1βˆ’y))dy         =βˆ’4∫_0 ^( 1) ln(y)dyβˆ’4li_2 (y)          =4βˆ’4ΞΆ(2)=4(1βˆ’ΞΆ(2))...       we know that:        1: li_2 (x)=βˆ’βˆ«_0 ^( x) ((ln(1βˆ’z))/z)dz=Ξ£_(n=1) ^∞ (x^n /n^2 )        1^βˆ— : li_2 (1)=ΞΆ(2)        1^(βˆ—βˆ—) :∫_0 ^( 1) ln(t)dt=βˆ’1
$$\:\:\:\:\:{solution}: \\ $$$$\:\:{y}^{\mathrm{2}} =\mathrm{1}βˆ’{x} \\ $$$$\:\:\boldsymbol{\phi}=\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\sqrt{\mathrm{1}βˆ’{x}}\:\right)}{\mathrm{1}βˆ’\sqrt{\mathrm{1}βˆ’{x}}}{dx} \\ $$$$\:\:\:\:\boldsymbol{\phi}=\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{yln}\left({y}\right)}{\mathrm{1}βˆ’{y}}{dy} \\ $$$$\:\:\:\:\:\:\:=βˆ’\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({y}\right){dy}βˆ’\mathrm{4}{li}_{\mathrm{2}} \left({y}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{4}βˆ’\mathrm{4}\zeta\left(\mathrm{2}\right)=\mathrm{4}\left(\mathrm{1}βˆ’\zeta\left(\mathrm{2}\right)\right)… \\ $$$$\:\:\:\:\:{we}\:{know}\:{that}: \\ $$$$\:\:\:\:\:\:\mathrm{1}:\:{li}_{\mathrm{2}} \left({x}\right)=βˆ’\int_{\mathrm{0}} ^{\:{x}} \frac{{ln}\left(\mathrm{1}βˆ’{z}\right)}{{z}}{dz}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\mathrm{1}^{\ast} :\:{li}_{\mathrm{2}} \left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\mathrm{1}^{\ast\ast} :\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}=βˆ’\mathrm{1} \\ $$$$\:\:\:\:\:\:\: \\ $$

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