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 )+....