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Question Number 125336 by mnjuly1970 last updated on 10/Dec/20
              ... nice  calculus ...    evaluate :     φ=∫_(0 ) ^( 1) x^2 ln(x).ln(1−x)dx =?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}\:… \\ $$$$\:\:{evaluate}\:: \\ $$$$\:\:\:\phi=\int_{\mathrm{0}\:} ^{\:\mathrm{1}} {x}^{\mathrm{2}} {ln}\left({x}\right).{ln}\left(\mathrm{1}−{x}\right){dx}\:=? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 10/Dec/20
−Σ_(n≥1) ^∞ (1/n)∫_0 ^1 x^(2+n) log(x)dx      log(x)=t  =−Σ_(n≥1) ^∞ (1/n)∫_(−∞) ^0 te^((n+3)t) dt      (n+3)t=−u  =−Σ_(n≥1) ^∞ (1/n)∫_0 ^∞ (u/((n+3)^2 ))e^(−u) du  =Σ^∞ (1/(n(n+3)^2 ))Γ(2)=(1/3)Σ^∞ (1/((n+3)))((1/n)−(1/((n+3))))  =(1/9)Σ^∞ (1/n)−(1/(n+3))−(1/3)Σ^∞ (1/((n+3)^2 ))  =(1/9)(1+(1/2)+(1/3))−(1/3)((π^2 /6)−1−(1/4)−(1/9))  =((11)/(54))+((49)/(108))−(π^2 /(18))=((71)/(108))−(π^2 /(18))
$$−\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}+{n}} {log}\left({x}\right){dx}\:\:\:\:\:\:{log}\left({x}\right)={t} \\ $$$$=−\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{−\infty} ^{\mathrm{0}} {te}^{\left({n}+\mathrm{3}\right){t}} {dt}\:\:\:\:\:\:\left({n}+\mathrm{3}\right){t}=−{u} \\ $$$$=−\underset{{n}\geqslant\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{u}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} }{e}^{−{u}} {du} \\ $$$$=\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{3}\right)^{\mathrm{2}} }\Gamma\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}}\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\overset{\infty} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{11}}{\mathrm{54}}+\frac{\mathrm{49}}{\mathrm{108}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}}=\frac{\mathrm{71}}{\mathrm{108}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$
Commented by mnjuly1970 last updated on 10/Dec/20
  very nice solution.  thanks alot...
$$\:\:{very}\:{nice}\:{solution}. \\ $$$${thanks}\:{alot}… \\ $$
Commented by Dwaipayan Shikari last updated on 10/Dec/20
With pleasure sir!
$${With}\:{pleasure}\:{sir}! \\ $$
Answered by Olaf last updated on 10/Dec/20
Let I_n  = ∫_0 ^1 x^n lnxdx, n∈N  I_n  = [(x^(n+1) /(n+1))lnx]_0 ^1 −∫_0 ^1 (x^(n+1) /(n+1)).(dx/x)  I_n  = −(1/(n+1))∫_0 ^1 x^n dx  I_n  = −(1/(n+1))[(x^(n+1) /(n+1))]_0 ^1  = −(1/((n+1)^2 ))  φ = ∫_0 ^1 x^2 lnxln(1−x)dx  φ = ∫_0 ^1 x^2 lnxΣ_(n=1) ^∞ (x^n /n)dx  φ = Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^(n+2) lnxdx  φ = Σ_(n=1) ^∞ (I_(n+2) /n)  φ = −Σ_(n=1) ^∞ (1/(n(n+3)^2 ))  φ = −(1/9)Σ_(n=1) ^∞ [(1/n)−(1/(n+3))−(3/((n+3)^2 ))]  φ = −(1/9)Σ_(n=1) ^∞ [(1/n)−(1/(n+3))]+(1/3)Σ_(n=1) ^∞ (1/((n+3)^2 ))  φ = −(1/9)[1+(1/2)+(1/3)]+(1/3)[Σ_(n=1) ^∞ (1/n^2 )−(1+(1/4)+(1/9))]  φ = −(1/9)[((11)/6)]+(1/3)[(π^2 /6)−((49)/(36))]  φ = (π^2 /(18))−((71)/(108))
$$\mathrm{Let}\:\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \mathrm{ln}{xdx},\:{n}\in\mathbb{N} \\ $$$$\mathrm{I}_{{n}} \:=\:\left[\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mathrm{ln}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}.\frac{{dx}}{{x}} \\ $$$$\mathrm{I}_{{n}} \:=\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {dx} \\ $$$$\mathrm{I}_{{n}} \:=\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\left[\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\phi\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$\phi\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \mathrm{ln}{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}{dx} \\ $$$$\phi\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{2}} \mathrm{ln}{xdx} \\ $$$$\phi\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{I}_{{n}+\mathrm{2}} }{{n}} \\ $$$$\phi\:=\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\phi\:=\:−\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}−\frac{\mathrm{3}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} }\right] \\ $$$$\phi\:=\:−\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right]+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\phi\:=\:−\frac{\mathrm{1}}{\mathrm{9}}\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right]+\frac{\mathrm{1}}{\mathrm{3}}\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}\right)\right] \\ $$$$\phi\:=\:−\frac{\mathrm{1}}{\mathrm{9}}\left[\frac{\mathrm{11}}{\mathrm{6}}\right]+\frac{\mathrm{1}}{\mathrm{3}}\left[\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{49}}{\mathrm{36}}\right] \\ $$$$\phi\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{18}}−\frac{\mathrm{71}}{\mathrm{108}} \\ $$
Commented by mnjuly1970 last updated on 10/Dec/20
grateful mr olaf...
$${grateful}\:{mr}\:{olaf}… \\ $$

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