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Question Number 157278 by mnjuly1970 last updated on 21/Oct/21
Answered by mindispower last updated on 21/Oct/21
tanh^− (x)=Σ_(m≥0) (x^(2m+1) /(2m+1)) ∫_0 ^1 ((ln(1−(√x))tanh^− ((√x)))/x)dx =∫_0 ^1 ((ln(1−(√x)))/( (√x)))tanh^− ((√x)).2d(√x) =∫_0 ^1 ((ln(1−y))/y)tanh^− (y)dy =2∫_0 ^1 Σ_(m≥0) (1/(2m+1))ln(1−y)y^(2m) dy =Σ_(m≥0) (2/(2m+1))∫_0 ^1 ln(1−y)y^(2m) dy =Σ_(m≥0) (2/(2m+1))∂_a ∫_0 ^1 (1−y)^a y^(2m) dy∣_(a=0) =Σ(2/(2m+1))∂_a β(2m+1,1+a)∣_(a=0) =Σ_(m≥0) (2/(2m+1))β(2m+1,1)(Ψ(1)−Ψ(2+2m)) =Σ_(m≥1) (2/(2m+1)).(1/(2m+1))(−H_(2m+1) ) =Σ_(m≥0) ((−2H_(2m+1) )/((2m+1)^2 ))
$${tanh}^{−} \left({x}\right)=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{m}+\mathrm{1}} }{\mathrm{2}{m}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\sqrt{{x}}\right){tanh}^{−} \left(\sqrt{{x}}\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\sqrt{{x}}\right)}{\:\sqrt{{x}}}{tanh}^{−} \left(\sqrt{{x}}\right).\mathrm{2}{d}\sqrt{{x}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{y}\right)}{{y}}\mathrm{tan}{h}^{−} \left({y}\right){dy} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{m}+\mathrm{1}}{ln}\left(\mathrm{1}−{y}\right){y}^{\mathrm{2}{m}} {dy} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}}{\mathrm{2}{m}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{y}\right){y}^{\mathrm{2}{m}} {dy} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}}{\mathrm{2}{m}+\mathrm{1}}\partial_{{a}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{y}\right)^{{a}} {y}^{\mathrm{2}{m}} {dy}\mid_{{a}=\mathrm{0}} \\ $$$$=\Sigma\frac{\mathrm{2}}{\mathrm{2}{m}+\mathrm{1}}\partial_{{a}} \beta\left(\mathrm{2}{m}+\mathrm{1},\mathrm{1}+{a}\right)\mid_{{a}=\mathrm{0}} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}}{\mathrm{2}{m}+\mathrm{1}}\beta\left(\mathrm{2}{m}+\mathrm{1},\mathrm{1}\right)\left(\Psi\left(\mathrm{1}\right)−\Psi\left(\mathrm{2}+\mathrm{2}{m}\right)\right) \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}}{\mathrm{2}{m}+\mathrm{1}}.\frac{\mathrm{1}}{\mathrm{2}{m}+\mathrm{1}}\left(−{H}_{\mathrm{2}{m}+\mathrm{1}} \right) \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{−\mathrm{2}{H}_{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 22/Oct/21
thank you so much sir power
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{power} \\ $$
Commented by mindispower last updated on 22/Oct/21
withe Pleasur
$${withe}\:{Pleasur} \\ $$
Answered by qaz last updated on 22/Oct/21
∫_0 ^1 ((ln(1−(√x))tanh^(−1) (√x))/x)dx =2∫_0 ^1 ((ln(1−x)tanh^(−1) x)/x)dx =2Σ_(n=0) ^∞ (1/(2n+1))∫_0 ^1 x^(2n) ln(1−x)dx =−2Σ_(n=0) ^∞ (H_(2n+1) /((2n+1)^2 )) =Σ_(n=1) ^∞ ((−1)^n −1)(H_n /n^2 ) −−−−−−−−−−−− H_n =∫_0 ^1 ((1−x^n )/(1−x))dx=−(1−x^n )ln(1−x)∣_0 ^1 −n∫_0 ^1 x^(n−1) ln(1−x)dx=−n∫_0 ^1 x^(n−1) ln(1−x)dx ⇒H_n =−n∫_0 ^1 x^(n−1) ln(1−x)dx
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\sqrt{\mathrm{x}}\right)\mathrm{tanh}^{−\mathrm{1}} \sqrt{\mathrm{x}}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{tanh}^{−\mathrm{1}} \mathrm{x}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{2}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2n}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$=−\mathrm{2}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\left(−\mathrm{1}\right)^{\mathrm{n}} −\mathrm{1}\right)\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$−−−−−−−−−−−− \\ $$$$\mathrm{H}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{n}} }{\mathrm{1}−\mathrm{x}}\mathrm{dx}=−\left(\mathrm{1}−\mathrm{x}^{\mathrm{n}} \right)\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{n}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}=−\mathrm{n}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$$$\Rightarrow\mathrm{H}_{\mathrm{n}} =−\mathrm{n}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx} \\ $$
Commented by mnjuly1970 last updated on 22/Oct/21
thanks alot sir qaz
$${thanks}\:{alot}\:{sir}\:{qaz} \\ $$
Answered by mnjuly1970 last updated on 22/Oct/21

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