Question

求无穷积分

Answer 1

∫[ln(1+1/x) - 1/(1+x) ]dx
=xln(1+1/x) - ∫[x /(1+1/x)](-1/x^2)dx - ln|1+x|
=xln(1+1/x) + ∫ x^2/(1+x) dx - ln|1+x|
=xln(1+1/x) + ∫ [x -1 + 1/(1+x)] dx - ln|1+x|
=xln(1+1/x) + (1/2)x^2 -x + C
∫(1->+∞) [ln(1+1/x) - 1/(1+x) ]dx
=[xln(1+1/x) + (1/2)x^2 -x] |(1->+∞)
=-(ln2+ 1/2-1) + lim(x->+∞) [ xln(1+1/x) + (1/2)x^2 -x ]
=1/2 - ln2 +lim(x->+∞) [ xln(1+1/x) + (1/2)x^2 -x ]
=1/2 - ln2 +0
=1/2 - ln2
//
let
y=1/x
y->0
ln(1+y) ~ y -(1/2)y^2 + (1/3)y^3
ln(1+y) +(1/2)(1/y)^2 -(1/y) ~ (1/3)y^3
lim(x->+∞) [ xln(1+1/x) + (1/2)x^2 -x ]
=lim(y->0) [ (1/y)ln(1+y) + (1/2)(1/y)^2 -(1/y) ]
=lim(y->0)(1/3)y^3
=0