nice-calculus-prove-that-0-1-ln-1-x-1-1-x-dx-4-1-2-

Question Number 135821 by mnjuly1970 last updated on 16/Mar/21

....nice ..... calculus.... prove that :: 𝛗=∫_0 ^( 1) (((ln(1−x))/(1−(√(1−x)))))dx=4(1−ζ(2))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:….{nice}\:\:\:…..\:\:\:{calculus}….\: \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}\right){dx}=\mathrm{4}\left(\mathrm{1}−\zeta\left(\mathrm{2}\right)\right) \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 16/Mar/21

Φ=∫_0 ^1 ((log(1−x))/(1−(√(1−x))))dx we do the changement(√(1−x))=t ⇒1−x=t^2 ⇒x=1−t^2 Φ =−∫_0 ^1 ((log(1−1+t^2 ))/(1−t))(−2t)dt =4∫_0 ^1 ((tlog(t))/(1−t))dt =4∫_0 ^1 tlogtΣ_(n=0) ^∞ t^n dt =4Σ_(n=0) ^∞ ∫_0 ^1 t^(n+1) logt dt =4Σ U_n U_n =∫_0 ^1 t^(n+1) logt dt =[(t^(n+2) /(n+2))logt]_0 ^1 −(1/(n+2))∫_0 ^1 t^(n+1) dt =−(1/((n+2)^2 )) ⇒Φ =−4 Σ_(n=0) ^∞ (1/((n+2)^2 )) =−4Σ_(n=2) ^∞ (1/n^2 ) =−4{Σ_(n=1) ^∞ (1/n^2 )−1} =4−4.(π^2 /6) =4−((2π^2 )/3)

$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\sqrt{\mathrm{1}−\mathrm{x}}=\mathrm{t}\:\Rightarrow\mathrm{1}−\mathrm{x}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\mathrm{1}−\mathrm{t}^{\mathrm{2}} \\ $$$$\Phi\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{log}\left(\mathrm{1}−\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{t}}\left(−\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{tlog}\left(\mathrm{t}\right)}{\mathrm{1}−\mathrm{t}}\mathrm{dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{tlogt}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{n}} \:\mathrm{dt}\:=\mathrm{4}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}+\mathrm{1}} \mathrm{logt}\:\mathrm{dt}\:=\mathrm{4}\Sigma\:\mathrm{U}_{\mathrm{n}} \\ $$$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}+\mathrm{1}} \mathrm{logt}\:\mathrm{dt}\:=\left[\frac{\mathrm{t}^{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{2}}\mathrm{logt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{t}^{\mathrm{n}+\mathrm{1}} \:\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }\:\Rightarrow\Phi\:=−\mathrm{4}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=−\mathrm{4}\sum_{\mathrm{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:=−\mathrm{4}\left\{\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }−\mathrm{1}\right\} \\ $$$$=\mathrm{4}−\mathrm{4}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\mathrm{4}−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$

Answered by Dwaipayan Shikari last updated on 16/Mar/21

∫_0 ^1 ((log(1−x))/(1−(√(1−x))))dx=∫_0 ^1 ((log(x))/(1−(√x)))dx =∫_0 ^1 ((4tlog(t))/(1−t))dt =4Φ′(2)=4(1−(π^2 /6)) Φ(α)=∫_0 ^1 ((1−x^(α−1) )/(1−x))dx=−γ+ψ(α) Φ′(α)=−∫_0 ^1 ((x^(α−1) log(x))/(1−x))dx=ψ^1 (α)⇒∫_0 ^1 ((xlog(x))/(1−x))=−ψ^1 (2) Φ′(2)=−Σ_(n=1) ^∞ (1/((n+1)^2 ))=−((π^2 /6)−1)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({x}\right)}{\mathrm{1}−\sqrt{{x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}{tlog}\left({t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$=\mathrm{4}\Phi'\left(\mathrm{2}\right)=\mathrm{4}\left(\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$\Phi\left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{\alpha−\mathrm{1}} }{\mathrm{1}−{x}}{dx}=−\gamma+\psi\left(\alpha\right) \\ $$$$\Phi'\left(\alpha\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\alpha−\mathrm{1}} {log}\left({x}\right)}{\mathrm{1}−{x}}{dx}=\psi^{\mathrm{1}} \left(\alpha\right)\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xlog}\left({x}\right)}{\mathrm{1}−{x}}=−\psi^{\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\Phi'\left(\mathrm{2}\right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=−\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right) \\ $$

Commented by mnjuly1970 last updated on 16/Mar/21

$${thanks}\:{alot}\:{mr}\:{payan}… \\ $$

Answered by Ñï= last updated on 16/Mar/21

∫_0 ^1 ((ln(1−x))/(1−(√(1−x))))dx=∫_0 ^1 ((lnx)/(1−(√x)))dx=∫_0 ^1 (((1−(√x))lnx)/(1−x))dx =∫_0 ^1 ((lnx)/(1−x))dx−∫_0 ^1 (((√x)lnx)/(1−x))dx =∫_0 ^1 ((lnx)/(1−x))dx−∫_0 ^1 ((4x^2 lnx)/(1−x^2 ))dx ((√x)→x) =−Li_2 (1)−4∫_0 ^1 (1−(1/(1−x^2 )))lnxdx =−Li_2 (1)−4∫_0 ^1 lnxdx+4∫_0 ^1 ((lnx)/((1−x)(1+x)))dx =−Li_2 (1)+4+2∫_0 ^1 ((1/(1−x))+(1/(1+x)))lnxdx =−Li_2 (1)+4−2Li_2 (1)+2Li_2 (−1) =−4Li_2 (1)+4 =4(1−ζ(2))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}−\sqrt{{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−\sqrt{{x}}\right){lnx}}{\mathrm{1}−{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}−{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{{x}}{lnx}}{\mathrm{1}−{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}−{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{4}{x}^{\mathrm{2}} {lnx}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\left(\sqrt{{x}}\rightarrow{x}\right) \\ $$$$=−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\right){lnxdx} \\ $$$$=−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {lnxdx}+\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{dx} \\ $$$$=−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{4}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){lnxdx} \\ $$$$=−{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{4}−\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{2}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$$=−\mathrm{4}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{4} \\ $$$$=\mathrm{4}\left(\mathrm{1}−\zeta\left(\mathrm{2}\right)\right) \\ $$

Commented by mnjuly1970 last updated on 16/Mar/21

$$\:\:{thank}\:{you}\:{do}\:{much}… \\ $$

Answered by mnjuly1970 last updated on 16/Mar/21

solution: y^2 =1−x 𝛗=2∫_0 ^( 1) ((ln((√(1−x)) ))/(1−(√(1−x))))dx 𝛗=4∫_0 ^( 1) ((yln(y))/(1−y))dy =−4∫_0 ^( 1) ln(y)dy−4li_2 (y) =4−4ζ(2)=4(1−ζ(2))... we know that: 1: li_2 (x)=−∫_0 ^( x) ((ln(1−z))/z)dz=Σ_(n=1) ^∞ (x^n /n^2 ) 1^∗ : li_2 (1)=ζ(2) 1^(∗∗) :∫_0 ^( 1) ln(t)dt=−1

$$\:\:\:\:\:{solution}: \\ $$$$\:\:{y}^{\mathrm{2}} =\mathrm{1}−{x} \\ $$$$\:\:\boldsymbol{\phi}=\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\sqrt{\mathrm{1}−{x}}\:\right)}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}{dx} \\ $$$$\:\:\:\:\boldsymbol{\phi}=\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{yln}\left({y}\right)}{\mathrm{1}−{y}}{dy} \\ $$$$\:\:\:\:\:\:\:=−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({y}\right){dy}−\mathrm{4}{li}_{\mathrm{2}} \left({y}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{4}−\mathrm{4}\zeta\left(\mathrm{2}\right)=\mathrm{4}\left(\mathrm{1}−\zeta\left(\mathrm{2}\right)\right)… \\ $$$$\:\:\:\:\:{we}\:{know}\:{that}: \\ $$$$\:\:\:\:\:\:\mathrm{1}:\:{li}_{\mathrm{2}} \left({x}\right)=−\int_{\mathrm{0}} ^{\:{x}} \frac{{ln}\left(\mathrm{1}−{z}\right)}{{z}}{dz}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\mathrm{1}^{\ast} :\:{li}_{\mathrm{2}} \left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\mathrm{1}^{\ast\ast} :\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\: \\ $$

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