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