Question

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

Answer 1

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`

Answer 2

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`