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Question Number 118924 by mnjuly1970 last updated on 21/Oct/20

... advanced calculus... prove that :: Σ_(n=1) ^∞ (((−1)^n H_n )/n^2 ) =∫_0 ^( 1) ((ln(1−x)ln(1+x) )/x)dx note :: H_n =Σ_(k=1) ^n (1/k) therefore: Σ_(n=1 ) ^∞ (((−1)^n H_n )/n^2 )=((−5)/8) ζ (3 ) ✓ ..m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}\right)\:\:}{{x}}{dx}\:\: \\ $$$$\:\:\:\:\:{note}\:::\:{H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$$\:\:\:\:\:\:\:{therefore}:\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} }=\frac{−\mathrm{5}}{\mathrm{8}}\:\zeta\:\left(\mathrm{3}\:\right)\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:..{m}.{n}.{july}.\mathrm{1970}... \\ $$$$ \\ $$

Answered by mindispower last updated on 21/Oct/20

∫_0 ^1 x^(n−1) ln(1−x)dx=−∫_0 ^1 x^(n−1) Σ_(k≥1) (x^k /k)dx =−Σ_k (1/k)∫_0 ^1 x^(n+k−1) dx=−Σ_(k≥1) (1/((n+k)k)) =−Σ_(k≥1) (1/n)((1/k)−(1/(n+k)))=−(1/n)Σ_(k≤n) (1/k)=−(H_n /n) ⇒∫_0 ^1 x^(n−1) ln(1−x)dx=−(H_n /n) (1/n)∫x^(n−1) ln(1−x)dx=−(H_n /n^2 ) ⇒Σ(((−1)^(n−1) )/n)∫_0 ^1 x^(n−1) ln(1−x)dx=Σ_(n≥1) (((−1)^n H_n )/n^2 ) ⇒∫_0 ^1 ((1/x)Σ(((−1)^(n−1) x^n )/n))ln(1−x)dx=Σ_(n≥1) (((−1)^n H_n )/n^2 ) Σ_(n≥1) (((−1)^(n−1) x^n )/n)=ln(1+x)⇔ ∫_0 ^1 ((ln(1+x)ln(1−x))/x)dx=Σ(((−1)^n H_n )/n^2 ) ∫_0 ^1 ((ln(1+x)ln(1−x))/x)dx done in your answer

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} \underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{k}} }{{k}}{dx} \\ $$$$=−\underset{{k}} {\sum}\frac{\mathrm{1}}{{k}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+{k}−\mathrm{1}} {dx}=−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}+{k}\right){k}} \\ $$$$=−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{n}+{k}}\right)=−\frac{\mathrm{1}}{{n}}\underset{{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{{k}}=−\frac{{H}_{{n}} }{{n}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}=−\frac{{H}_{{n}} }{{n}} \\ $$$$\frac{\mathrm{1}}{{n}}\int{x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}=−\frac{{H}_{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\Sigma\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}−{x}\right){dx}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}\Sigma\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}} }{{n}}\right){ln}\left(\mathrm{1}−{x}\right){dx}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}} }{{n}}={ln}\left(\mathrm{1}+{x}\right)\Leftrightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}\:{done}\:{in}\:{your}\:{answer} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 21/Oct/20

bravo bravo mr power grateful sir...

$${bravo}\:{bravo} \\ $$$${mr}\:{power}\: \\ $$$$\:{grateful}\:{sir}... \\ $$

Commented by mindispower last updated on 21/Oct/20

withe pleasur

$${withe}\:{pleasur} \\ $$$$ \\ $$

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