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