Question Number 114541 by mnjuly1970 last updated on 19/Sep/20

Answered by abdomsup last updated on 19/Sep/20

A=∫_0 ^1 ((x^2 ln(x))/(1−x^2 ))dx ⇒ A =∫_0 ^1 x^2 ln(x)Σ_(n=0) ^∞ x^(2n) dx =Σ_(n=0) ^∞ ∫_0 ^1 x^(2n+2) ln(x)dx u_n =∫_0 ^(1 ) x^(2n+2) ln(x)dx =[(x^(2n+3) /(2n+3))ln(x)]_0 ^1 −∫_0 ^1 (x^(2n+2) /(2n+3))dx =−(1/((2n+3)^2 ))[x^(2n+3) ]_0 ^1 =−(1/((2n+3)^2 )) ⇒A =−Σ_(n=0) ^(∞ ) (1/((2n+3)^2 )) =_(n=p−1) −Σ_(p=1) ^(∞ ) (1/((2p+1)^2 )) we have Σ_(p=1) ^∞ (1/p^2 ) =(1/4)Σ_(p=1) ^∞ (1/p^2 ) +Σ_(p=0) ^∞ (1/((2p+1)^2 )) ⇒ Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(3/4)×(π^2 /6) =(π^2 /8) ⇒Σ_(p=1) ^∞ (1/((2p+1)^2 )) =(π^2 /8)−1 ⇒ A =1−(π^2 /8)

Commented bymnjuly1970 last updated on 19/Sep/20

thank you .very nice..

Commented bymathmax by abdo last updated on 19/Sep/20

you are welcome