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Question Number 144787 by mnjuly1970 last updated on 29/Jun/21

Q :: # Calculus # If : 𝛗 ( n ) : = ∫_0 ^( 1) (( x^( 2n) )/(1 + x^( 2) )) dx then find the value of :: S := Σ_(n=1) ^∞ ((( −1 )^( n) 𝛗 ( n ))/n) = ? m.n.july.1970

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Q}\:::\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:#\:\mathrm{Calculus}\:# \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{If}\::\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:\left(\:{n}\:\right)\::\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}^{\:\mathrm{2}{n}} }{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:\mathrm{d}{x}\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{then}\:\:\mathrm{find}\:\:\mathrm{the}\:\:\mathrm{value}\:\mathrm{of}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\:−\mathrm{1}\:\right)^{\:{n}} \:\boldsymbol{\phi}\:\left(\:{n}\:\right)}{{n}}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{m}.\mathrm{n}.\mathrm{july}.\mathrm{1970} \\ $$

Answered by qaz last updated on 29/Jun/21

S=Σ_(n=1) ^∞ (((−1)^n )/n)φ(n) =∫_0 ^1 Σ_(n=1) ^∞ (((−1)^n )/n)∙(x^(2n) /(1+x^2 ))dx =∫_0 ^1 ((ln(1+x^2 ))/(1+x^2 ))dx =∫_0 ^∞ ((ln(1+x^2 ))/(1+x^2 ))dx−∫_1 ^∞ ((ln(1+x^2 ))/(1+x^2 ))dx =−2∫_0 ^(π/2) lncos udu−∫_0 ^1 ((ln(1+x^2 )−2lnx)/(1+x^2 ))dx =πln2−S−2G ⇒S=(π/2)ln2−G

$$\mathrm{S}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\phi\left(\mathrm{n}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\centerdot\frac{\mathrm{x}^{\mathrm{2n}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{lncos}\:\mathrm{udu}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\pi\mathrm{ln2}−\mathrm{S}−\mathrm{2G} \\ $$$$\Rightarrow\mathrm{S}=\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\mathrm{G} \\ $$

Commented by mnjuly1970 last updated on 29/Jun/21

thx mr qaz...

$${thx}\:\:{mr}\:{qaz}... \\ $$

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