Question Number 101793 by Dwaipayan Shikari last updated on 04/Jul/20

∫_1 ^2 ((logu)/(((√(u−1)))((√(u−1))+1)))du

Answered by mathmax by abdo last updated on 05/Jul/20

I =∫_1 ^(2 )  ((ln(u))/((√(u−1))((√(u−1))+1)))du  changement (√(u−1))=x give u−1 =x^2  ⇒  I =∫_0 ^1   ((ln(1+x^2 ))/(x(x+1))) (2x)dx =2 ∫_0 ^1  ((ln(1+x^2 ))/(1+x)) dx  =2 ∫_0 ^1  ln(1+x^2 )Σ_(n=0) ^∞ (−1)^n  x^n  =2 Σ_(n=0) ^∞  (−1)^n  ∫_0 ^1  x^n  ln(1+x^2 )dx  =2 Σ_(n=0) ^∞  (−1)^n  A_n   with A_n =∫_0 ^1  x^n ln(1+x^2 )dx  by parts we get  A_n =[(x^(n+1) /(n+1))ln(1+x^2 )]_0 ^1  −∫_0 ^1  (x^(n+1) /(n+1))×((2x)/(1+x^2 ))dx  =((ln(2))/(n+1))−(2/(n+1))∫_0 ^1   (x^(n+2) /(1+x^2 ))dx   ∫_0 ^1  (x^(n+2) /(1+x^2 ))dx =∫_0 ^1  (((x^2 +1−1)x^n )/(1+x^2 ))dx =∫_0 ^1  x^n  dx−∫_0 ^1  (x^n /(x^2  +1))dx  =(1/(n+1))−∫_0 ^(1 )  (x^n /(x^2  +1))dx  we have  ∫_0 ^1  (x^n /(x^2  +1))dx =∫_0 ^1  x^n (Σ_(k=0) ^∞  (−1)^k  x^(2k) )  =Σ_(k=0) ^∞  (−1)^k  ∫_0 ^1  x^(n+2k)  dx =Σ_(k=0) ^∞  (−1)^k  ×(1/(n+2k+1)) ⇒  A_n =((ln2)/(n+1))−(2/(n+1)){(1/(n+1))−Σ_(k=0) ^∞  (((−1)^k )/(2k+n+1))}  =((ln2)/(n+1))−(2/((n+1)^2 )) −(2/(n+1)) Σ_(k=0) ^∞  (((−1)^k )/(2k+n+1)) ⇒  I =2ln2 Σ_(n=0) ^∞  (((−1)^n )/(n+1))−4 Σ_(n=0) ^∞  (((−1)^n )/((n+1)^2 )) −4 Σ_(n=0) ^∞  (((−1)^n )/(n+1))(Σ_(k=0) ^∞  (((−1)^k )/(2k+n+1)))  Σ_(n=0) ^∞  (((−1)^n )/(n+1)) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n) =ln(2)  Σ_(n=0) ^∞  (((−1)^n )/((n+1)^2 )) =Σ_(n=1) ^∞  (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞  (((−1)^n )/n^2 ) =−{2^(1−2) −1)ξ(2)  =−(−(1/2))×(π^2 /6) =(π^2 /(12))  Σ_(n=0) ^∞  (((−1)^n )/(n+1))Σ_(k=0) ^∞  (((−1)^k )/(2k+n+1)) =Σ_(n=0) ^∞  Σ_(k=0) ^∞  (((−1)^(n+k) )/((n+1)(2k+n+1)))  ...be continued...

Commented by1549442205 last updated on 06/Jul/20

Thank you sir.Please,can you show me  about the function ξ(n) that used in solution?