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Question Number 151230 by mnjuly1970 last updated on 19/Aug/21

prove: ∫_0 ^( ∞) (( ln ( 1+x^( 2) ))/(x^( 2) (1+x^( 2) )))dx= π ln((e/2) ) ..

$$ \\ $$$$\:\:\:\:{prove}: \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}\:\left(\:\mathrm{1}+{x}^{\:\mathrm{2}} \right)}{{x}^{\:\mathrm{2}} \left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)}{dx}=\:\pi\:{ln}\left(\frac{{e}}{\mathrm{2}}\:\right)\:.. \\ $$

Commented by Lordose last updated on 19/Aug/21

Commented by Lordose last updated on 19/Aug/21

Answered by Lordose last updated on 19/Aug/21

Ω = ∫_0 ^( 1) ((ln(1+x^2 ))/(x^2 (1+x^2 )))dx =^(P.F) ∫_0 ^( 1) ((ln(1+x^2 ))/x^2 )dx − ∫_0 ^( 1) ((ln(1+x^2 ))/(1+x^2 ))dx Ω = Φ − Ψ Φ =^(IBP) −((ln(1+x^2 ))/x)∣_0^+ ^1 + ∫_0 ^( 1) (2/(1+x^2 ))dx Φ = −log(2) + (𝛑/2) Ψ = ∫_0 ^( 1) ((ln(1+x^2 ))/(1+x^2 ))dx =^(x=tan(y)) −2∫_0 ^(𝛑/4) ln(cos(y))dy Ψ = 2∫_0 ^(𝛑/4) (Σ_(k=1) ^∞ (((−1)^k )/k)cos(2ky) + log(2))dy = 2Σ_(k=1) ^∞ (((−1)^k )/k)∫_0 ^(𝛑/4) cos(2ky)dy + 2log(2)∫_0 ^(𝛑/4) 1dy Ψ = Σ_(k=1) ^∞ (((−1)^k sin(((𝛑k)/2)))/k^2 ) + ((𝛑log(2))/2) Ψ = (𝛑/2)log(2) − G Ω = (𝛑/2) − log(2) − (𝛑/2)log(2) + G 𝛀 = (𝛑/2) − log(2)(1 + (𝛑/2)) + G G = Catalan′s Constant = Σ_(k=1) ^∞ (((−1)^k sin(((𝛑k)/2)))/k^2 )

$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:\overset{\boldsymbol{\mathrm{P}}.\boldsymbol{\mathrm{F}}} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:−\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\Omega\:=\:\Phi\:−\:\Psi \\ $$$$\Phi\:\overset{\boldsymbol{\mathrm{IBP}}} {=}−\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mid_{\mathrm{0}^{+} } ^{\mathrm{1}} \:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\Phi\:=\:−\mathrm{log}\left(\mathrm{2}\right)\:+\:\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\Psi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\overset{\mathrm{x}=\mathrm{tan}\left(\mathrm{y}\right)} {=}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\left(\mathrm{y}\right)\right)\mathrm{dy} \\ $$$$\Psi\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\mathrm{cos}\left(\mathrm{2ky}\right)\:+\:\mathrm{log}\left(\mathrm{2}\right)\right)\mathrm{dy}\:=\:\mathrm{2}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{k}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \mathrm{cos}\left(\mathrm{2ky}\right)\mathrm{dy}\:+\:\mathrm{2log}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \mathrm{1dy} \\ $$$$\Psi\:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{sin}\left(\frac{\boldsymbol{\pi}\mathrm{k}}{\mathrm{2}}\right)}{\mathrm{k}^{\mathrm{2}} }\:+\:\frac{\boldsymbol{\pi}\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Psi\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)\:−\:\boldsymbol{\mathrm{G}} \\ $$$$\Omega\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:−\:\mathrm{log}\left(\mathrm{2}\right)\:−\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)\:+\:\boldsymbol{\mathrm{G}} \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:−\:\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)\left(\mathrm{1}\:+\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)\:+\:\boldsymbol{\mathrm{G}} \\ $$$$\boldsymbol{\mathrm{G}}\:=\:\mathrm{Catalan}'\mathrm{s}\:\mathrm{Constant}\:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{sin}\left(\frac{\boldsymbol{\pi}\mathrm{k}}{\mathrm{2}}\right)}{\mathrm{k}^{\mathrm{2}} } \\ $$

Commented by mnjuly1970 last updated on 19/Aug/21

thank you so much mr lordose my mistake Ω=∫_0 ^( ∞) ((ln(1+x^( 2) ))/(x^2 (1+x^( 2) )))dx grateful...

$$\:\:\:{thank}\:{you}\:{so}\:{much}\:{mr}\:{lordose} \\ $$$$\:\:\:{my}\:{mistake}\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)}{dx} \\ $$$$\:\:\:\:\:\:{grateful}... \\ $$

Answered by qaz last updated on 19/Aug/21

∫_0 ^1 ((ln(1+x^2 ))/(x^2 (1+x^2 )))dx =∫_0 ^1 ((ln(1+x^2 ))/x^2 )dx−∫_0 ^1 ((ln(1+x^2 ))/(1+x^2 ))dx =−((ln(1+x^2 ))/x)∣_0 ^1 +∫_0 ^1 (2/(1+x^2 ))dx+2∫_0 ^(π/4) lncos xdx =−ln2+(π/2)+∫_0 ^(π/2) lncos (x/2)dx =−ln2+(π/2)+(1/2)∫_0 ^(π/2) ln((1+cos x)/2)dx =−ln2+(π/2)+(1/2)∫_0 ^(π/2) ln((sin x)/(tan (x/2)))dx−(1/2)∫_0 ^(π/2) ln2dx =−ln2+(π/2)−(π/2)ln2−∫_0 ^(π/4) lntan xdx =−ln2+(π/2)−(π/2)ln2−∫_0 ^1 ((lnx)/(1+x^2 ))dx =−ln2+(π/2)−(π/2)ln2−Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) lnxdx =−ln2+(π/2)−(π/2)ln2−Σ_(n=0) ^∞ (((−1)^(n+1) )/((2n+1)^2 )) =−ln2+(π/2)−(π/2)ln2+G

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lncos}\:\mathrm{xdx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}+\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{lncos}\:\frac{\mathrm{x}}{\mathrm{2}}\mathrm{dx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{2}}\mathrm{dx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln}\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}}\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{ln2dx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{lntan}\:\mathrm{xdx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2n}} \mathrm{lnxdx} \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\mathrm{ln2}+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\mathrm{ln2}+\mathrm{G} \\ $$

Answered by Lordose last updated on 19/Aug/21

Ω = ∫_0 ^( ∞) ((ln(1+x^2 ))/(x^2 (1+x^2 )))dx =^(x=tan(y)) −2∫_0 ^(𝛑/2) cot^2 yln(cos(y))dy Ω = −2(∂/∂a)∣_(a=2) ∫_0 ^(𝛑/2) cos^a (y)sin^(−2) (y)dy Ω = −(∂/∂a)∣_(a=2) (B(((a+1)/2),−(1/2))) = −(1/2)(𝛙^((0)) (((a+1)/2))−𝛙^((0)) ((a/2)))B(((a+1)/2),−(1/2)) 𝛀 = (1/2)𝛑(𝛄 + 𝛙^((0)) ((3/2)))

$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:\overset{\mathrm{x}=\mathrm{tan}\left(\mathrm{y}\right)} {=}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cot}^{\mathrm{2}} \mathrm{yln}\left(\mathrm{cos}\left(\mathrm{y}\right)\right)\mathrm{dy} \\ $$$$\Omega\:=\:−\mathrm{2}\frac{\partial}{\partial\mathrm{a}}\mid_{\mathrm{a}=\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{cos}^{\mathrm{a}} \left(\mathrm{y}\right)\mathrm{sin}^{−\mathrm{2}} \left(\mathrm{y}\right)\mathrm{dy} \\ $$$$\Omega\:=\:−\frac{\partial}{\partial\mathrm{a}}\mid_{\mathrm{a}=\mathrm{2}} \left(\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)−\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\frac{\mathrm{a}}{\mathrm{2}}\right)\right)\boldsymbol{\mathrm{B}}\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}},−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\pi}\left(\boldsymbol{\gamma}\:+\:\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\ $$

Answered by qaz last updated on 19/Aug/21

∫_0 ^∞ ((ln(1+x^2 ))/(x^2 (1+x^2 )))dx =∫_0 ^(π/2) ((lnsec^2 x)/(tan^2 x))dx =−2∫_0 ^(π/2) cot^2 xlncos xdx =2∫_0 ^(π/2) (1−csc^2 x)lncos xdx =−πln2−2∫_0 ^(π/2) csc^2 xlncos xdx =−πln2+2cot xlncos x∣_0 ^(π/2) +2∫_0 ^(π/2) dx =−πln2+π

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{lnsec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cot}\:^{\mathrm{2}} \mathrm{xlncos}\:\mathrm{xdx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\mathrm{1}−\mathrm{csc}^{\mathrm{2}} \mathrm{x}\right)\mathrm{lncos}\:\mathrm{xdx} \\ $$$$=−\pi\mathrm{ln2}−\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{csc}^{\mathrm{2}} \mathrm{xlncos}\:\mathrm{xdx} \\ $$$$=−\pi\mathrm{ln2}+\mathrm{2cot}\:\mathrm{xlncos}\:\mathrm{x}\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} +\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{dx} \\ $$$$=−\pi\mathrm{ln2}+\pi \\ $$

Commented by mnjuly1970 last updated on 19/Aug/21

thanks alot..

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

Commented by peter frank last updated on 19/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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