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Question Number 28247 by abdo imad last updated on 22/Jan/18
find the value of ∫_1 ^(+∞) ((lnx)/(1+x^2 ))dx .
findthevalueof∫1+∞lnx1+x2dx.
Commented by abdo imad last updated on 24/Jan/18
let put I= ∫_1 ^(+∞) ((lnx)/(1+x^2 ))dx I= −∫_0 ^1 ((lnx)/(1+x^2 ))dx + ∫_0 ^∞ ((lnx)/(1+x^2 ))dx but we have proved that ∫_0 ^∞ ((lnx)/(1+x^2 ))dx =0 so I= −∫_0 ^1 ((lnx)/(1+x^2 ))dx = −∫_0 ^1 (Σ_(n=0) ^(+∞) (−1)^n x^(2n) )lnxdx =−Σ_(n=0) ^∝ (−1)^n ∫_0 ^1 x^(2n) lnx dx and by parts ∫_0 ^1 x^(2n) lnx dx= [ (1/(2n+1))x^(2n+1) lnx]_0 ^1 − ∫_0 ^1 (1/(2n+1)) x^(2n) dx = −(1/((2n+1)^2 )) so I= Σ_(n=0) ^∝ (((−1)^n )/((2n+1)^2 )) thissum is known be continued........
letputI=∫1+∞lnx1+x2dxI=−∫01lnx1+x2dx+∫0∞lnx1+x2dxbutwehaveprovedthat∫0∞lnx1+x2dx=0soI=−∫01lnx1+x2dx=−∫01(∑n=0+∞(−1)nx2n)lnxdx=−∑n=0∝(−1)n∫01x2nlnxdxandbyparts∫01x2nlnxdx=[12n+1x2n+1lnx]01−∫0112n+1x2ndx=−1(2n+1)2soI=∑n=0∝(−1)n(2n+1)2thissumisknownbecontinued……..

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