Question Number 210098 by klipto last updated on 30/Jul/24
$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{1}}\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by AlagaIbile last updated on 30/Jul/24
![(1/(x^(2 ) − 1)) = - Σ_(n=0) ^∞ x^(2n) ⇒ -Σ_(n=0) ^∞ ∫_0 ^( 1) x^(2n) ln x dx = Σ_(n=0) ^∞ (1/((2n + 1)^2 )) = (1/1^2 ) + (1/3^2 ) + (1/5^2 ) + (1/7^2 ) + ... = [(1/1^2 ) + (1/2^2 ) + (1/3^2 ) + (1/(4^2 )) +... ] − (1/2^2 )[(1/1^2 ) + (1/2^2 ) + (1/3^2 ) + (1/4^2 ) + ...] = (3/4) [(1/1^2 ) + (1/2^2 ) + (1/3^2 ) + (1/4^2 ) + ∙∙∙] = (3/4) (π^2 /6) = (π^2 /8) ■](https://www.tinkutara.com/question/Q210099.png)
$$\:\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} \:−\:\mathrm{1}}\:=\:-\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{x}^{\mathrm{2}{n}} \\ $$$$\:\Rightarrow\:-\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\mathrm{2}{n}} \:\mathrm{ln}\:{x}\:{dx}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }\:+\:… \\ $$$$\:=\:\left[\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} \:}\:+…\:\right]\:−\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\:+\:…\right] \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\left[\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\:+\:\centerdot\centerdot\centerdot\right] \\ $$$$\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{great}}\:\boldsymbol{\mathrm{bigman}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{chat}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{whatapp}} \\ $$
Commented by AlagaIbile last updated on 02/Aug/24
$$\:{No}\:{problem}.\:+\mathrm{2348131291501} \\ $$
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{alrt}}\:\boldsymbol{\mathrm{now}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{man}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{go}}\:\boldsymbol{\mathrm{message}}\:\boldsymbol{\mathrm{you}} \\ $$
Answered by Spillover last updated on 01/Aug/24
Answered by Spillover last updated on 02/Aug/24
Answered by Spillover last updated on 02/Aug/24
Answered by Spillover last updated on 02/Aug/24
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{chat}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{whatsapp}} \\ $$
Commented by Spillover last updated on 02/Aug/24
$${no} \\ $$
Answered by Spillover last updated on 02/Aug/24
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{chat}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{whatsapp}}? \\ $$
Commented by Spillover last updated on 02/Aug/24
$${no} \\ $$
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{about}}\:\boldsymbol{\mathrm{facebook}}\: \\ $$
Commented by Spillover last updated on 02/Aug/24
$${no} \\ $$
Commented by klipto last updated on 02/Aug/24
$$\boldsymbol{\mathrm{okay}} \\ $$