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Question Number 61661 by maxmathsup by imad last updated on 05/Jun/19

1) calculate ∫∫_R^+^2 ((dxdy)/((1+x^2 )(1+y^2 ))) 2) find the value of ∫_0 ^∞ ((ln(x))/(x^2 −1)) dx .

1)calculate∫∫R+2dxdy(1+x2)(1+y2)2)findthevalueof∫0∞ln(x)x2−1dx.

Commented by maxmathsup by imad last updated on 07/Jun/19

1) ∫∫_R^+^2 ((dxdy)/((1+x^2 )(1+y^2 ))) =∫_0 ^∞ (dx/(1+x^2 )) .∫_0 ^∞ (dy/(1+y^2 )) =(π/2).(π/2) =(π^2 /4) 2) let A =∫_0 ^∞ ((ln(x))/(x^2 −1)) ⇒−A =∫_0 ^∞ ((ln(x))/(1−x^2 ))dx =∫_0 ^1 ((ln(x))/(1−x^2 ))dx +∫_1 ^(+∞) ((ln(x))/(1−x^2 )) dx ∫_0 ^1 ((ln(x))/(1−x^2 ))dx =∫_0 ^1 lnx(Σ_(n=0) ^∞ x^(2n) )dx =Σ_(n=0) ^∞ ∫_0 ^1 x^(2n) ln(x)dx by parts ∫_0 ^1 x^(2n) ln(x)dx =[(1/(2n+1))x^(2n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/((2n+1)))x^(2n) dx =−(1/((2n+1)^2 )) ⇒∫_0 ^1 ((ln(x))/(1−x^2 ))dx =−Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4)(π^2 /6) =(π^2 /8) ⇒ ∫_0 ^1 ((ln(x))/(1−x^2 ))dx =−(π^2 /8) ∫_1 ^(+∞) ((ln(x))/(1−x^2 )) dx =_(x =(1/t)) − ∫_0 ^1 ((−ln(t))/(1−(1/t^2 ))) (−(dt/t^2 )) =−∫_0 ^1 ((ln(t))/(t^2 −1))dt =∫_0 ^1 ((ln(t))/(1−t^2 )) dt =−(π^2 /8) ⇒ −A =−(π^2 /8)−(π^2 /8) ⇒A =(π^2 /4) .

1)∫∫R+2dxdy(1+x2)(1+y2)=∫0∞dx1+x2.∫0∞dy1+y2=π2.π2=π242)letA=∫0∞ln(x)x2−1⇒−A=∫0∞ln(x)1−x2dx=∫01ln(x)1−x2dx+∫1+∞ln(x)1−x2dx∫01ln(x)1−x2dx=∫01lnx(∑n=0∞x2n)dx=∑n=0∞∫01x2nln(x)dxbyparts∫01x2nln(x)dx=[12n+1x2n+1ln(x)]01−∫011(2n+1)x2ndx=−1(2n+1)2⇒∫01ln(x)1−x2dx=−∑n=0∞1(2n+1)2∑n=1∞1n2=14∑n=1∞1n2+∑n=0∞1(2n+1)2⇒∑n=0∞1(2n+1)2=34π26=π28⇒∫01ln(x)1−x2dx=−π28∫1+∞ln(x)1−x2dx=x=1t−∫01−ln(t)1−1t2(−dtt2)=−∫01ln(t)t2−1dt=∫01ln(t)1−t2dt=−π28⇒−A=−π28−π28⇒A=π24.

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