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Question Number 138145 by mnjuly1970 last updated on 10/Apr/21

...........advanced ... ... ... calculus......... find the value of:: Θ=Σ_(n=1) ^∞ (((−1)^n H_(2n) )/n)=???

$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...........{advanced}\:...\:...\:...\:{calculus}......... \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Theta=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{H}_{\mathrm{2}{n}} }{{n}}=??? \\ $$

Answered by Dwaipayan Shikari last updated on 10/Apr/21

Σ_(n=1) ^∞ H_(2n) x^(2n) =H_2 x^2 +H_4 x^4 +H_6 x^6 +H_8 x^8 +...=ψ(x^2 ) And Σ_(n=1) ^∞ (((−1)^n H_(2n) )/n)=∫_0 ^1 ((ψ(−x))/( (√x)))dx Σ_(n=1) ^∞ H_n x^n =H_1 x+H_2 x^2 +H_3 x^3 +...=−((log(1−x))/(1−x)) Σ_(n=1) ^∞ (−1)^n H_n x^n =−H_1 x+H_2 x^2 −H_3 x^3 +...=−((log(1+x))/(1+x)) ψ(x^2 )=−(1/2)(((log(1−x))/(1−x))+((log(1+x))/(1+x))) ψ(x)=−(1/2)(((log(1−(√x)))/(1−(√x)))+((log(1+(√x)))/(1+(√x)))) ∫_0 ^1 ((ψ(−x))/( (√x)))=−(1/2)∫_0 ^1 (1/( i(√x)))(((log(1−i(√x)))/(1−i(√x)))+((log(1+i(√x)))/(1+i(√x))))dx x=t^2 =−(1/(2i))∫_0 ^1 ((log(1−it))/(1−it))+((log(1+it))/(1+it))dt =−(1/(2i))∫_0 ^1 ((log(1+t^2 ))/(1+t^2 ))+itlog(((1−it)/(1+it)))dt ....

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{\mathrm{2}{n}} {x}^{\mathrm{2}{n}} ={H}_{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{4}} {x}^{\mathrm{4}} +{H}_{\mathrm{6}} {x}^{\mathrm{6}} +{H}_{\mathrm{8}} {x}^{\mathrm{8}} +...=\psi\left({x}^{\mathrm{2}} \right) \\ $$$${And}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{\mathrm{2}{n}} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\psi\left(−{x}\right)}{\:\sqrt{{x}}}{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} ={H}_{\mathrm{1}} {x}+{H}_{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{3}} {x}^{\mathrm{3}} +...=−\frac{{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} {x}^{{n}} =−{H}_{\mathrm{1}} {x}+{H}_{\mathrm{2}} {x}^{\mathrm{2}} −{H}_{\mathrm{3}} {x}^{\mathrm{3}} +...=−\frac{{log}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}} \\ $$$$\psi\left({x}^{\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}+\frac{{log}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}\right) \\ $$$$\psi\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{log}\left(\mathrm{1}−\sqrt{{x}}\right)}{\mathrm{1}−\sqrt{{x}}}+\frac{{log}\left(\mathrm{1}+\sqrt{{x}}\right)}{\mathrm{1}+\sqrt{{x}}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\psi\left(−{x}\right)}{\:\sqrt{{x}}}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:{i}\sqrt{{x}}}\left(\frac{{log}\left(\mathrm{1}−{i}\sqrt{{x}}\right)}{\mathrm{1}−{i}\sqrt{{x}}}+\frac{{log}\left(\mathrm{1}+{i}\sqrt{{x}}\right)}{\mathrm{1}+{i}\sqrt{{x}}}\right){dx}\:\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{it}\right)}{\mathrm{1}−{it}}+\frac{{log}\left(\mathrm{1}+{it}\right)}{\mathrm{1}+{it}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }+{itlog}\left(\frac{\mathrm{1}−{it}}{\mathrm{1}+{it}}\right){dt} \\ $$$$.... \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 10/Apr/21

grateful for your attention and effort...

$${grateful}\:\:{for}\:{your}\:{attention}\: \\ $$$$\:{and}\:\:{effort}...\: \\ $$

Answered by Ñï= last updated on 10/Apr/21

Σ_(n=1) ^∞ (((−1)^n )/n)H_(2n) =−∫_0 ^1 (1/(1−u))Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n (1−u^(2n) )du x=1 =∫_0 ^1 (1/(u−1))Σ_(n=1) ^∞ (((−1)^(n−1) )/n)[x^n −(xu^2 )^n ]du =∫_0 ^1 (1/(u−1))[ln(1+x)−ln(1+xu^2 )]du =.....???

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{H}_{\mathrm{2}{n}} =−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{u}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \left(\mathrm{1}−{u}^{\mathrm{2}{n}} \right){du}\:\:\:\:\:\:\:\:\:{x}=\mathrm{1} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{u}−\mathrm{1}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\left[{x}^{{n}} −\left({xu}^{\mathrm{2}} \right)^{{n}} \right]{du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{u}−\mathrm{1}}\left[{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}+{xu}^{\mathrm{2}} \right)\right]{du} \\ $$$$=.....??? \\ $$

Commented by mnjuly1970 last updated on 10/Apr/21

thanks alot mr...

$$\:\:\:\:\:\:{thanks}\:{alot}\:{mr}... \\ $$

Answered by mnjuly1970 last updated on 10/Apr/21

f(x):=Σ_(n=1 ) ^∞ (H_n /n) x^n =(1/2)ln^2 (1−x)+li_2 (x) f (i):=Σ(H_n /n)(i)^n =(1/2)ln^2 (1−i)+li_2 (i)....(∗) f(−i):=Σ_(n=1) ^∞ (H_n /n)(−i)^n =(1/2)ln^2 (1+i)+li_2 (−i)....(∗∗) (∗)+(∗∗) : Σ(H_n /n)(i^n +(−i)^n )=(1/2){ln^2 ((√2) e^((−iπ)/4) )+ln^2 ((√2) e^((iπ)/4) )}+(1/2)li_2 (i^2 ) ∴ Σ_(n=1) ^∞ (H_(2n) /(2n))((−1)^n +(−1)^n )=(1/2){(ln((√2) )−((iπ)/4))^2 +(ln((√(2 )) ) +((iπ)/4))^2 }−(π^2 /(24)) =(1/2){2((1/4)ln^2 (2)−(π^2 /(16)))}−(π^2 /(24)) ∴ Θ =(1/4)ln^2 (2)−((5π^2 )/(48)) ...✓✓

$$\:\:\:\:\:\:{f}\left({x}\right):=\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{{n}} }{{n}}\:{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{2}} \left({x}\right) \\ $$$$\:\:\:\:\:\:{f}\:\left({i}\right):=\Sigma\frac{\mathrm{H}_{{n}} }{{n}}\left({i}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{i}\right)+{li}_{\mathrm{2}} \left({i}\right)....\left(\ast\right) \\ $$$$\:\:\:\:\:\:\:\:{f}\left(−{i}\right):=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{{n}} }{{n}}\left(−{i}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+{i}\right)+{li}_{\mathrm{2}} \left(−{i}\right)....\left(\ast\ast\right) \\ $$$$\:\:\:\:\:\left(\ast\right)+\left(\ast\ast\right)\::\:\Sigma\frac{\mathrm{H}_{{n}} }{{n}}\left({i}^{{n}} +\left(−{i}\right)^{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}\:{e}^{\frac{−{i}\pi}{\mathrm{4}}} \right)+{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right\}+\frac{\mathrm{1}}{\mathrm{2}}{li}_{\mathrm{2}} \left({i}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\therefore\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{\mathrm{2}{n}} }{\mathrm{2}{n}}\left(\left(−\mathrm{1}\right)^{{n}} +\left(−\mathrm{1}\right)^{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({ln}\left(\sqrt{\mathrm{2}}\:\right)−\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} +\left({ln}\left(\sqrt{\mathrm{2}\:\:}\:\right)\:+\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} \right\}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\right)\right\}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\:\:\therefore\:\:\:\:\:\:\Theta\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}\:...\checkmark\checkmark \\ $$

Answered by Kamel last updated on 10/Apr/21

Commented by Kamel last updated on 10/Apr/21

Without complex numbers.

$${Without}\:{complex}\:{numbers}. \\ $$

Commented by mnjuly1970 last updated on 10/Apr/21

thanks alot..

$${thanks}\:{alot}.. \\ $$$$\: \\ $$

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