How do you show whether the improper integral #int ln(x)/x^3 dx# converges or diverges from 1 to infinity?

2 Answers
Apr 23, 2018

Converges:
#int_1^oolnx/x^3=1/4#

Explanation:

First, let's determine the indefinite integral #int(lnx)/x^3dx#

We'll use Integration by Parts, making the following selections:

#u=lnx#

#du=x^-1dx#

#dv=x^-3dx#

#v=-1/2x^-2#

#uv-intvdu=-lnx/(2x^2)+1/2intx^-2x^-1dx#

#=-lnx/(2x^2)+1/2intx^-3dx=-lnx/(2x^2)-1/(4x^2)#

Now that we have the antiderivative, we can get to evaluating the improper integral:

#int_1^oolnx/x^3dx=lim_(t->oo)int_1^tlnx/x^3dx#

#=lim_(t->oo)(-lnx/(2x^2)-1/(4x^2))|_1^t#

#=lim_(t->oo)(-lnt/(2t^2)-1/(4t^2)+ln1/2+1/4)#

#lim_(t->oo)-lnt/(2t^2)=-oo/oo# -- Indeterminate, we'll have to use l'Hospital's Rule:

#=lim_(t->oo)-(1/t)/(4t)=lim_(t->oo)-1/(4t)^3=0#

And,

#lim_(t->oo)-1/(4t^2)=0#

#ln1/2=0#

Then, we're left only with the #1/4# and the integral converges to

#int_1^oolnx/x^3=1/4#

Apr 23, 2018

See below

Explanation:

#int_1^(oo) ln(x)/x^3 dx#

#= int_1^(oo) ln(x) \ d(- 1/(2x^2)) #

By IBP:

#= ( - ln(x) *1/(2x^2) )_1^(oo) +1/2 int_1^( oo) d(ln(x)) \ 1/(x^2) #

#= 0 + 1/2 int_1^(oo) \ 1/(x^3) \ dx#

#= ( - 1/(4x^2) )_1^(oo) = 1/4#