Question Number 103343 by Dwaipayan Shikari last updated on 14/Jul/20

∫_0 ^1 x^(−x) dx

Answered by mathmax by abdo last updated on 15/Jul/20

at form of serie   I =∫_0 ^1  x^(−x)  dx =∫_0 ^1  e^(−xlnx) dx =∫_0 ^1  (Σ_(n=0) ^∞  (((−xlnx)^n )/(n!)))dx  =Σ_(n=0) ^∞  (((−1)^n )/(n!)) ∫_0 ^1  x^n  (lnx)^n  dx let A_n =∫_0 ^1  x^n (lnx)^n  dx  lnx =−t ⇒A_n =−∫_0 ^∞   (e^(−t) )^n  (−t)^n  (−e^(−t) )dt  =(−1)^n  ∫_0 ^∞  e^(−(n+1)t)  t^n  dt  =_((n+1)t =u)    (−1)^n  ∫_0 ^∞ e^(−u)  (u^n /((n+1)^n ))×(du/(n+1))  =(((−1)^n )/((n+1)^(n+1) )) ∫_0 ^∞  u^n  e^(−u)  du =(((−1)^n  Γ(n+1))/((n+1)^(n+1) )) =(((−1)^n n!)/((n+1)^(n+1) )) ⇒  I =Σ_(n=0) ^∞  (((−1)^n )/(n!))×(((−1)^n n!)/((n+1)^(n+1) )) =Σ_(n=0) ^∞  (1/((n+1)^(n+1) )) =Σ_(n=1) ^∞  (1/n^n )  =(1/1^1 ) +(1/2^2 ) +(1/3^3 )+....