Question Number 138145 by mnjuly1970 last updated on 10/Apr/21
$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:………..{advanced}\:…\:…\:…\:{calculus}……… \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Theta=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{H}_{\mathrm{2}{n}} }{{n}}=??? \\ $$
Answered by Dwaipayan Shikari last updated on 10/Apr/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{\mathrm{2}{n}} {x}^{\mathrm{2}{n}} ={H}_{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{4}} {x}^{\mathrm{4}} +{H}_{\mathrm{6}} {x}^{\mathrm{6}} +{H}_{\mathrm{8}} {x}^{\mathrm{8}} +…=\psi\left({x}^{\mathrm{2}} \right) \\ $$$${And}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {H}_{\mathrm{2}{n}} }{{n}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\psi\left(−{x}\right)}{\:\sqrt{{x}}}{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{H}_{{n}} {x}^{{n}} ={H}_{\mathrm{1}} {x}+{H}_{\mathrm{2}} {x}^{\mathrm{2}} +{H}_{\mathrm{3}} {x}^{\mathrm{3}} +…=−\frac{{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {H}_{{n}} {x}^{{n}} =−{H}_{\mathrm{1}} {x}+{H}_{\mathrm{2}} {x}^{\mathrm{2}} −{H}_{\mathrm{3}} {x}^{\mathrm{3}} +…=−\frac{{log}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}} \\ $$$$\psi\left({x}^{\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{log}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}+\frac{{log}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}\right) \\ $$$$\psi\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{log}\left(\mathrm{1}−\sqrt{{x}}\right)}{\mathrm{1}−\sqrt{{x}}}+\frac{{log}\left(\mathrm{1}+\sqrt{{x}}\right)}{\mathrm{1}+\sqrt{{x}}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\psi\left(−{x}\right)}{\:\sqrt{{x}}}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:{i}\sqrt{{x}}}\left(\frac{{log}\left(\mathrm{1}−{i}\sqrt{{x}}\right)}{\mathrm{1}−{i}\sqrt{{x}}}+\frac{{log}\left(\mathrm{1}+{i}\sqrt{{x}}\right)}{\mathrm{1}+{i}\sqrt{{x}}}\right){dx}\:\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{it}\right)}{\mathrm{1}−{it}}+\frac{{log}\left(\mathrm{1}+{it}\right)}{\mathrm{1}+{it}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }+{itlog}\left(\frac{\mathrm{1}−{it}}{\mathrm{1}+{it}}\right){dt} \\ $$$$…. \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 10/Apr/21
$${grateful}\:\:{for}\:{your}\:{attention}\: \\ $$$$\:{and}\:\:{effort}…\: \\ $$
Answered by Ñï= last updated on 10/Apr/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{H}_{\mathrm{2}{n}} =−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{u}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \left(\mathrm{1}−{u}^{\mathrm{2}{n}} \right){du}\:\:\:\:\:\:\:\:\:{x}=\mathrm{1} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{u}−\mathrm{1}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\left[{x}^{{n}} −\left({xu}^{\mathrm{2}} \right)^{{n}} \right]{du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{u}−\mathrm{1}}\left[{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}+{xu}^{\mathrm{2}} \right)\right]{du} \\ $$$$=…..??? \\ $$
Commented by mnjuly1970 last updated on 10/Apr/21
$$\:\:\:\:\:\:{thanks}\:{alot}\:{mr}… \\ $$
Answered by mnjuly1970 last updated on 10/Apr/21
$$\:\:\:\:\:\:{f}\left({x}\right):=\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{{n}} }{{n}}\:{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{li}_{\mathrm{2}} \left({x}\right) \\ $$$$\:\:\:\:\:\:{f}\:\left({i}\right):=\Sigma\frac{\mathrm{H}_{{n}} }{{n}}\left({i}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{i}\right)+{li}_{\mathrm{2}} \left({i}\right)….\left(\ast\right) \\ $$$$\:\:\:\:\:\:\:\:{f}\left(−{i}\right):=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{{n}} }{{n}}\left(−{i}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left(\mathrm{1}+{i}\right)+{li}_{\mathrm{2}} \left(−{i}\right)….\left(\ast\ast\right) \\ $$$$\:\:\:\:\:\left(\ast\right)+\left(\ast\ast\right)\::\:\Sigma\frac{\mathrm{H}_{{n}} }{{n}}\left({i}^{{n}} +\left(−{i}\right)^{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}\:{e}^{\frac{−{i}\pi}{\mathrm{4}}} \right)+{ln}^{\mathrm{2}} \left(\sqrt{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right\}+\frac{\mathrm{1}}{\mathrm{2}}{li}_{\mathrm{2}} \left({i}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\therefore\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{H}_{\mathrm{2}{n}} }{\mathrm{2}{n}}\left(\left(−\mathrm{1}\right)^{{n}} +\left(−\mathrm{1}\right)^{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({ln}\left(\sqrt{\mathrm{2}}\:\right)−\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} +\left({ln}\left(\sqrt{\mathrm{2}\:\:}\:\right)\:+\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} \right\}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\right)\right\}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$\:\:\therefore\:\:\:\:\:\:\Theta\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}\:…\checkmark\checkmark \\ $$
Answered by Kamel last updated on 10/Apr/21
Commented by Kamel last updated on 10/Apr/21
$${Without}\:{complex}\:{numbers}. \\ $$
Commented by mnjuly1970 last updated on 10/Apr/21
$${thanks}\:{alot}.. \\ $$$$\: \\ $$