Question

Improper integrals ? `int_e^oo(1)/(x*(ln x)^3)dx`

`int_e^oo(1)/(x*(ln x)^3)dx`

Answer 1

`1/2`

Explanation:

First find `int 1/(x (lnx ) ^3) dx`

`= int 1/x * 1/(lnx)^3 dx`

Let `u = lnx`

`du = 1/x dx`

`=> int 1/u^3 du`

`=> int u^(-3) du`

`= -1/2 u^(-2) + c`

`= -1/(2 (lnx)^2 ) + c`

`int_e ^oo 1/(x (lnx ) ^3) dx = lim_(alpha to oo ) int_e ^ alpha 1/(x (lnx ) ^3) dx`

`=> lim_(alpha to oo) [ -1/( 2 (lnx)^2 ) ]_e ^ alpha`

`=> lim_(alpha to oo) {-1/(2(lna)^2 ) - ( -1/2 )}`

as `alpha to oo, ln(alpha) to oo` `=> 1/(ln(a) ) to 0`

`= 1/2`

Answer 2

The answer is `=1/2`

Explanation:

First, calculate the indefinite integral by substitution

Let `u=lnx`, `=>`, `du=(dx)/x`

Then, the integral is

`I=int(dx)/(x ln^3x)=int(du)/(u^3)`

`=-1/(2u^2)`

`=-1/(2ln^2x)+C`

Therefore, the improper integral is

`lim_(p->+oo)int_e ^p(dx)/(x ln^3x)`

`=lim_(p->+oo)[-1/(2ln^2x)]_e ^ p`

`=lim_(p->+oo)((-1/(2ln^2p))-(-1/(2ln^2e)))`

`=lim_(p->+oo)(1/2-1/(2ln^2p))`

`=1/2`