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Question Number 128922 by mnjuly1970 last updated on 11/Jan/21
...advanced calculus... evaluate ::: Ω=∫_(0 ) ^( (1/2)) ((Arctanh(x))/x)dx=?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:\:{evaluate}\::::\: \\ $$$$\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}\:} ^{\:\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{Arctanh}\left({x}\right)}{{x}}{dx}=? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 11/Jan/21
tanh^(−1) (x)=y ⇒x=((e^y −e^(−y) )/(e^y +e^(−y) ))⇒xe^(2y) +x=e^(2y) −1⇒e^(2y) =((1+x)/(1−x)) y=(1/2)log(((1+x)/(1−x))) (1/2)∫_0 ^(1/2) ((log(1+x))/x)−((log(1−x))/x)dx =(1/2)Σ_(n=1) ^∞ (−1)^(n+1) (1/n)∫_0 ^(1/2) x^(n−1) dx +(1/2)Σ_(n=1) ^∞ (1/n)∫_0 ^(1/2) x^(n−1) dx =(1/2)Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 )((1/2))^n +(1/2)Σ_(n=1) ^∞ (1/n^2 )((1/2))^n =Σ_(n=0) ^∞ (1/((2n+1)^2 ))((1/2))^(2n+1) =(1/8)Σ_(n=0) ^∞ (1/((n+(1/2))^2 ))((1/4))^n =(1/8)Σ_(n=0) ^∞ (((1)_n )/((n+(1/2))^2 n!))((1/4))^n
$${tanh}^{−\mathrm{1}} \left({x}\right)={y} \\ $$$$\Rightarrow{x}=\frac{{e}^{{y}} −{e}^{−{y}} }{{e}^{{y}} +{e}^{−{y}} }\Rightarrow{xe}^{\mathrm{2}{y}} +{x}={e}^{\mathrm{2}{y}} −\mathrm{1}\Rightarrow{e}^{\mathrm{2}{y}} =\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{log}\left(\mathrm{1}+{x}\right)}{{x}}−\frac{{log}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{{n}−\mathrm{1}} {dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {x}^{{n}−\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}\right)_{{n}} }{\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} {n}!}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} \:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Dwaipayan Shikari last updated on 11/Jan/21
I think it can be fransformed into Hypergeometric functions Trying to find that..
$${I}\:{think}\:{it}\:{can}\:{be}\:{fransformed}\:{into}\:{Hypergeometric}\:{functions} \\ $$$${Trying}\:{to}\:{find}\:{that}.. \\ $$
Commented by mnjuly1970 last updated on 11/Jan/21
thank you so much.. li_2 (x) =−∫_0 ^( x) ((ln(1−t))/t)dt li_2 (x)+li_2 (1−x)=(π^2 /6) −ln(x)ln(1−x) of course you know these better than that me. i just mentioned it...(dilogarithm function properties)..
$${thank}\:{you}\:{so}\:{much}.. \\ $$$$\:\:{li}_{\mathrm{2}} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\:{x}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$$\:\:\:{li}_{\mathrm{2}} \left({x}\right)+{li}_{\mathrm{2}} \left(\mathrm{1}−{x}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right) \\ $$$$\:\:{of}\:{course}\:{you}\:{know} \\ $$$$\:\:{these}\:{better}\:{than}\:{that}\:{me}. \\ $$$$\:\:{i}\:{just}\:{mentioned}\:{it}…\left({dilogarithm}\:{function}\:{properties}\right).. \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 11/Jan/21
Or −(1/2)Σ_(n=1) ^∞ (1/n^2 )(−(1/2))^n +(1/2)Σ_(n=1) ^∞ (1/n^2 )((1/2))^n =−(1/2)Li_2 (−(1/2))+(1/2)Li_2 ((1/2))
$${Or}\: \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Commented by mnjuly1970 last updated on 11/Jan/21
very nice thank you mr dwaipayan
$${very}\:{nice}\: \\ $$$${thank}\:{you}\:{mr}\:{dwaipayan} \\ $$

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