>>403
>>408
f(t) = log((1+t)^(-1-t)) - log((1+2t+(3/2)t^2+t^3/2+t^4/24)e^(-3t))
と置くと
f''(t) = -(t^3/(1+t))(192+456t+384t^2+140t^3+20t^4+t^5))/(24+48t+36t^2+12t^3+t^4)^2
≦ 0 (t≧0)
であり、t≧0でf'(t),f(t)は単調減少で0以下より
∫[0,∞](1+t)^(-1-t)dt
≦ ∫[0,∞](1+2t+(3/2)t^2+t^3/2+t^4/24)e^(-3t)dt
= 1/3+2/9+1/9+1/27+1/243

一方
∫[0,1]x^(-x)dx
= Σ[n=1,∞]n^(-n)
< 1+1/2^2+1/3^3+1/4^4+(1/5^5)/(1-1/5)

ゆえに
∫[0,∞]x^(-x)dx
< 1+1/2^2+1/3^3+1/4^4+(1/5^5)/(1-1/5) + 1/3+2/9+1/9+1/27+1/243
= 77727427/38880000
< 2