Question Number 100054 by  M±th+et+s last updated on 24/Jun/20

I_(n,m) =∫_0 ^1 ∫_0 ^1 (((ln(x))^n (ln(y))^m )/(1−xy))dx dy

Answered by maths mind last updated on 24/Jun/20

since ∣xy∣<1  (1/(1−xy))=Σ_(k≥0) (xy)^k   I =∫∫((ln^n (x)ln^m (y))/(1−xy))dxdy=∫_0 ^1 ∫_0 ^1 Σ_(k≥0) (x^k y^k )ln^n (x)ln^m (y)dxdy  =Σ_(k≥0) ∫_0 ^1 ∫_0 ^1 (x^k ln^n (x)dx)y^k ln^m (y)dy  f(s)=∫_0 ^1 x^k ln^s (x)dx=∫_0 ^(+∞) (−t)^s e^(−(k+1)t) dt  t→(k+1)t⇔  f(s)=(−1)^s ∫_0 ^(+∞) ((t^s e^(−t) )/((k+1)^(s+1) ))dt=(−1)^s ((Γ(s+1))/((k+1)^(s+1) ))  I=Σ_(k≥0) ∫_0 ^1 x^k ln^n (x)dx.∫_0 ^1 y^k ln^m (x)dx  =Σ_(k≥0) (((−1)^n Γ(n+1))/((k+1)^(n+1) )).(−1)^m ((Γ(m+1))/((k+1)^(m+1) ))  =(−1)^(n+m) Γ(n+1)Γ(m+1).Σ_(k≥0) (1/((1+k)^(n+m+2) ))  =(−1)^(n+m) n!.m!.ζ(n+m+2)  ⇔I_(n,m) =∫_0 ^1 ∫_0 ^1 ((ln^n (x)ln^m (y))/(1−xy))dxdy=(−1)^(n+m) n!.m!ζ(n+m+2)

Commented by M±th+et+s last updated on 24/Jun/20

cool sir. thank you

Commented bymaths mind last updated on 24/Jun/20

withe pleasur