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

Apr 23, 2018

Converges:
${\int}_{1}^{\infty} \ln \frac{x}{x} ^ 3 = \frac{1}{4}$

#### Explanation:

First, let's determine the indefinite integral $\int \frac{\ln x}{x} ^ 3 \mathrm{dx}$

We'll use Integration by Parts, making the following selections:

$u = \ln x$

$\mathrm{du} = {x}^{-} 1 \mathrm{dx}$

$\mathrm{dv} = {x}^{-} 3 \mathrm{dx}$

$v = - \frac{1}{2} {x}^{-} 2$

$u v - \int v \mathrm{du} = - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \int {x}^{-} 2 {x}^{-} 1 \mathrm{dx}$

$= - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \int {x}^{-} 3 \mathrm{dx} = - \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}}$

Now that we have the antiderivative, we can get to evaluating the improper integral:

${\int}_{1}^{\infty} \ln \frac{x}{x} ^ 3 \mathrm{dx} = {\lim}_{t \to \infty} {\int}_{1}^{t} \ln \frac{x}{x} ^ 3 \mathrm{dx}$

$= {\lim}_{t \to \infty} \left(- \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}}\right) {|}_{1}^{t}$

$= {\lim}_{t \to \infty} \left(- \ln \frac{t}{2 {t}^{2}} - \frac{1}{4 {t}^{2}} + \ln \frac{1}{2} + \frac{1}{4}\right)$

${\lim}_{t \to \infty} - \ln \frac{t}{2 {t}^{2}} = - \frac{\infty}{\infty}$ -- Indeterminate, we'll have to use l'Hospital's Rule:

$= {\lim}_{t \to \infty} - \frac{\frac{1}{t}}{4 t} = {\lim}_{t \to \infty} - \frac{1}{4 t} ^ 3 = 0$

And,

${\lim}_{t \to \infty} - \frac{1}{4 {t}^{2}} = 0$

$\ln \frac{1}{2} = 0$

Then, we're left only with the $\frac{1}{4}$ and the integral converges to

${\int}_{1}^{\infty} \ln \frac{x}{x} ^ 3 = \frac{1}{4}$

Apr 23, 2018

See below

#### Explanation:

${\int}_{1}^{\infty} \ln \frac{x}{x} ^ 3 \mathrm{dx}$

$= {\int}_{1}^{\infty} \ln \left(x\right) \setminus d \left(- \frac{1}{2 {x}^{2}}\right)$

By IBP:

$= {\left(- \ln \left(x\right) \cdot \frac{1}{2 {x}^{2}}\right)}_{1}^{\infty} + \frac{1}{2} {\int}_{1}^{\infty} d \left(\ln \left(x\right)\right) \setminus \frac{1}{{x}^{2}}$

$= 0 + \frac{1}{2} {\int}_{1}^{\infty} \setminus \frac{1}{{x}^{3}} \setminus \mathrm{dx}$

$= {\left(- \frac{1}{4 {x}^{2}}\right)}_{1}^{\infty} = \frac{1}{4}$