Menu Close

prove-that-0-1-ln-1-x-1-x-dx-x-1-x-2-pi-2-2-m-n-

Tinku Tara 0 Comments
Pin




Question Number 164103 by mnjuly1970 last updated on 14/Jan/22
prove that Ω=∫_0 ^( 1) ln(((1+x)/(1−x)) ).(dx/(x (√( 1−x^( 2) )))) = (π^( 2) /2) −− m.n−−
$$ \\ $$$$\:\:\:{prove}\:{that} \\ $$$$\: \\ $$$$\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\:\right).\frac{{dx}}{{x}\:\sqrt{\:\mathrm{1}−{x}^{\:\mathrm{2}} }}\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:−−\:{m}.{n}−− \\ $$$$ \\ $$
Answered by Lordose last updated on 14/Jan/22
Ω =^(x=((1−x)/(1+x))) 2∫_0 ^( 1) ((ln((1/x)))/((1+x)^2 (((1−x)/(1+x))(√(1−(((1−x)/(1+x)))^2 ))))dx Ω = ∫_0 ^( 1) ((ln((1/x)))/( (√x)(1−x)))dx =^(x=x^2 ) 4∫_0 ^( 1) ((ln((1/x)))/((1−x^2 )))dx Ω = −4Σ_(k=1) ^∞ ∫_0 ^( 1) x^(2k) ln(x)dx =^(IBP) 4Σ_(k=1) ^∞ (1/((2k+1)^2 )) Ω = Σ_(k=1) ^∞ (1/((k+(1/2))^2 ))dx = 𝛙^((1)) ((1/2)) 𝛀 = (𝛑^2 /2) ▲▲▲
$$\Omega\:\overset{\mathrm{x}=\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}} {=}\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)^{\mathrm{2}} }\right.}\mathrm{dx} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}−\mathrm{x}\right)}\mathrm{dx}\:\overset{\mathrm{x}=\mathrm{x}^{\mathrm{2}} } {=}\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$\Omega\:=\:−\mathrm{4}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{x}^{\mathrm{2k}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\:\overset{\boldsymbol{\mathrm{IBP}}} {=}\mathrm{4}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Omega\:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\mathrm{dx}\:=\:\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{\Omega}\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{2}}\:\blacktriangle\blacktriangle\blacktriangle \\ $$
Commented by mnjuly1970 last updated on 14/Jan/22
mercey
$${mercey} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *