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

Jun 13, 2018

${\int}_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = + \infty$

#### Explanation:

As the function is continuous in the closed intervals $x \in \left[\frac{1}{2} , 1\right)$ and $x \in \left(1 , 2\right]$ evaluate:

${\int}_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = {\int}_{\frac{1}{2}}^{1} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} + {\int}_{1}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}}$

Change $t = \frac{1}{x}$ in the second integral:

${\int}_{1}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = {\int}_{1}^{\frac{1}{2}} - \frac{t}{t} ^ 2 \frac{\mathrm{dt}}{\ln \left(\frac{1}{t}\right)} ^ 4$

exchange the limits of integration:

${\int}_{1}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = {\int}_{\frac{1}{2}}^{1} \frac{\mathrm{dt}}{t {\left(- \ln \left(t\right)\right)}^{4}}$

${\int}_{1}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = {\int}_{\frac{1}{2}}^{1} \frac{\mathrm{dt}}{t {\left(\ln \left(t\right)\right)}^{4}}$

Then:

${\int}_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = 2 {\int}_{\frac{1}{2}}^{1} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}}$

${\int}_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = 2 {\int}_{\frac{1}{2}}^{1} \frac{d \left(\ln x\right)}{\ln x} ^ 4$

${\int}_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = {\left[- \frac{2}{3 {\left(\ln x\right)}^{3}}\right]}_{\frac{1}{2}}^{1}$

${\int}_{\frac{1}{2}}^{2} \frac{\mathrm{dx}}{x {\left(\ln x\right)}^{4}} = \frac{2}{3 {\left(\ln \left(\frac{1}{2}\right)\right)}^{3}} - \frac{2}{3} {\lim}_{x \to {1}^{-}} \frac{1}{\ln x} ^ 3 = + \infty$