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Question Number 124723 by mnjuly1970 last updated on 05/Dec/20
.... nice calculus.... prove that:: challanging integral:: Ω= ∫_1 ^( ∞) ((({x}−(1/2))/x))dx=^(???) ln((√(2π)) )−1 {x} is fractional part of ′x′
….nicecalculus….provethat::challangingintegral::Ω=∫1∞({x}−12x)dx=???ln(2π)−1{x}isfractionalpartof′x′
Answered by Bird last updated on 05/Dec/20
Ω=∫_1 ^∞ ((x−[x]−(1/2))/x)dx =∫_1 ^∞ (1−(([x]+(1/2))/x))dx =Σ_(n=1) ^∞ ∫_n ^(n+1) (1−((n+(1/2))/x))dx =Σ_(n=1) ^∞ ∫_n ^(n+1) dx−(n+(1/2))(dx/x) =Σ_(n=1) ^∞ (1−(n+(1/2)){ln(n+1)−ln(n)} =Σ_(n=1) ^∞ (1−(n+(1/2))ln(1+(1/n))) ln^′ (1+u)=1−u+o(u^2 ) ⇒ ln(1+u)=u−(u^2 /2) +o(u^2 ) ⇒ ln(1+(1/n))=(1/n)−(1/(2n^2 ))+o((1/n^2 ))⇒ 1−(n+(1/2))ln(1+(1/n)) ∼1−(n+(1/2))((1/n)−(1/(2n^2 ))) 1−(1−(1/(2n))+(1/(2n))−(1/(4n^2 )))=(1/(4n^2 )) ⇒ Σ_(n=1) ^∞ {1−(n+(1/2))ln(1+(1/n))} is convergent...be continued
Ω=∫1∞x−[x]−12xdx=∫1∞(1−[x]+12x)dx=∑n=1∞∫nn+1(1−n+12x)dx=∑n=1∞∫nn+1dx−(n+12)dxx=∑n=1∞(1−(n+12){ln(n+1)−ln(n)}=∑n=1∞(1−(n+12)ln(1+1n))ln′(1+u)=1−u+o(u2)⇒ln(1+u)=u−u22+o(u2)⇒ln(1+1n)=1n−12n2+o(1n2)⇒1−(n+12)ln(1+1n)∼1−(n+12)(1n−12n2)1−(1−12n+12n−14n2)=14n2⇒∑n=1∞{1−(n+12)ln(1+1n)}isconvergent…becontinued

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