Does this integral converge or do not converge? int_(1/2)^2(1)/(x(lnx)^4)dx

1 Answer
Jun 13, 2018

int_(1/2)^2 dx/(x(lnx)^4) = +oo

Explanation:

As the function is continuous in the closed intervals x in[1/2,1) and x in (1,2] evaluate:

int_(1/2)^2 dx/(x(lnx)^4) = int_(1/2)^1 dx/(x(lnx)^4) + int_1^2 dx/(x(lnx)^4)

Change t= 1/x in the second integral:

int_1^2 dx/(x(lnx)^4) = int_1^(1/2) -t/t^2 dt/(ln(1/t))^4

exchange the limits of integration:

int_1^2 dx/(x(lnx)^4) = int_(1/2)^1 dt/(t(-ln(t))^4)

int_1^2 dx/(x(lnx)^4) = int_(1/2)^1 dt/(t(ln(t))^4)

Then:

int_(1/2)^2 dx/(x(lnx)^4) = 2int_(1/2)^1 dx/(x(lnx)^4)

int_(1/2)^2 dx/(x(lnx)^4) = 2int_(1/2)^1 (d(lnx))/(lnx)^4

int_(1/2)^2 dx/(x(lnx)^4) = [-2/(3(lnx)^3)]_(1/2)^1

int_(1/2)^2 dx/(x(lnx)^4) = 2/(3(ln(1/2))^3)-2/3 lim_(x->1^-) 1/(lnx)^3 = +oo