# Prove that Integral \int \frac { d x } { x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } } = -...

## Question:

Prove that Integral

{eq}\int \frac { d x } { x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } } = - 6 ( \operatorname { log } _ { e } x ) ^ { - 1 / 6 } {/eq}

and hence determine whether the series {eq}\sum _ { n = 2 } ^ { \infty } \frac { 1 } { n ( \operatorname { log } _ { e } n ) ^ { 7 / 6 } } {/eq} is convergent or divergent.

## Series Integral Test for Convergence:

Firstly, we'll simplify the given indefinite integral using the general substitution method of integrals shown below:

• {eq}\displaystyle \int f'(x)f(x)\ dx=\int u\ du ,\ u= f(x) {/eq}

For the convergence of the given series, we'll use the integral test of series by replacing {eq}n {/eq} by {eq}x {/eq} of the expression of summation notation and computing the integral of the obtained function with the given limits.

The given indefinite integral with the result is:

{eq}\displaystyle \int \frac { d x } { x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } } = - 6 ( \operatorname { log } _ { e } x ) ^ { - 1 / 6 } {/eq}

Here, we have:

{eq}I=\displaystyle \int \frac { d x } { x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } } {/eq}

For the above expression, we'll take {eq}\operatorname { log } _ { e } x {/eq} as {eq}u {/eq}.

Differentiating the substituting with their respective variables, we get:

{eq}\displaystyle \ du=\frac{1}{x}\ dx {/eq}

Plug the above values of the substitution in the indefinite integral expression, we get:

{eq}\begin{align*} \displaystyle I&=\displaystyle \int \frac { 1} { u ^ { 7 / 6 } } \ du\\ &=\displaystyle \int u ^ { -7 / 6 } \ du&\because \frac{1}{u^n}=u^{-n}\\ &=\displaystyle \frac{u ^ { -7 / 6 +1}}{-7 / 6 +1}+C\\ &=\displaystyle \frac{u ^ { -1 / 6 }}{-1 / 6 }+C\\ &=\displaystyle -6u ^ { -1 / 6 }+C\\ \end{align*} {/eq}

Where,

• {eq}C {/eq} is the constant of integration.

Back substitute the value of the variable {eq}u {/eq} in the obtained integral expression.

{eq}\begin{align*} \displaystyle I&=\displaystyle -6(\operatorname { log } _ { e } x)^ { -1 / 6 }+C\\ \displaystyle \int \frac { d x } { x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } }&=\displaystyle -6(\operatorname { log } _ { e } x)^ { -1 / 6 }+C\\ \end{align*} {/eq}

Hence, it proved.

The given summation notation expression is:

{eq}\displaystyle \sum _ { n = 2 } ^ { \infty } \frac { 1 } { n ( \operatorname { log } _ { e } n ) ^ { 7 / 6 } } {/eq}

According to the series integral test, we'll replace {eq}n {/eq} by {eq}x {/eq} of the above expression.

{eq}\displaystyle f(x)= \frac { 1 } {x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } } {/eq}

For the convergence of the series by the integral test, we have:

{eq}\displaystyle \int_{2}^{\infty }\frac { 1 } {x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } }\ dx=\lim_{t \to\infty }\int_{2}^{t }\frac { 1 } {x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } }\ dx {/eq}

From part (a), we have:

{eq}\displaystyle \int \frac { d x } { x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } } = - 6 ( \operatorname { log } _ { e } x ) ^ { - 1 / 6 } {/eq}

Now, the integral expression of the summation notation is:

{eq}\begin{align*} \displaystyle \int_{2}^{\infty }\frac { 1 } {x ( \operatorname { log } _ { e } x ) ^ { 7 / 6 } }\ dx&=\lim_{t \to\infty }\left [ - 6 ( \operatorname { log } _ { e } x ) ^ { - 1 / 6 } \right ]_{2}^{t }\\ &=\lim_{t \to\infty }\left [ - 6 ( \operatorname { log } _ { e } t ) ^ { - 1 / 6 }-(- 6 ( \operatorname { log } _ { e } 2 ) ^ { - 1 / 6 }) \right ]\\ &=\left [ - 6 ( \operatorname { log } _ { e } \infty ) ^ { - 1 / 6 }+ 6 ( \operatorname { log } _ { e } 2 ) ^ { - 1 / 6 } \right ]\\ &=\infty\\ &=\textrm{Undefined} \end{align*} {/eq}

The value of the integral is undefined so the given series is divergent.