Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (f)/Solution 1

To use the integral test, we need to compute

${\begin{aligned}\int _{2}^{\infty }{\frac {1}{x(\ln x)^{p}}}dx\end{aligned}}$

Let $u=\ln x$, then $du={\frac {1}{x}}dx$. With this change of variables, when $x=2$ then $u=\ln 2$ and when $x=\infty$ then $u$ does as well. Applying the substitution,

${\begin{aligned}\int _{2}^{\infty }{\frac {1}{x(\ln x)^{p}}}dx=\int _{\ln 2}^{\infty }{\frac {1}{u^{p}}}du.\end{aligned}}$

Any restrictions on $p$ will come if we cannot substitute the infinite upper limit and get zero. We have the following for the indefinite integral of $u^{-p}$:

$\int {\frac {1}{u^{p}}}{\textrm {d}}u={\begin{cases}{\frac {1}{1-p}}{\frac {1}{u^{p-1}}},&p>1\\\ln u,&p=1\\{\frac {1}{1-p}}u^{1-p},&p<1.\end{cases}}$

The only case in which plugging in and infinite limit will result in zero is for $p>1$. Since the Right hand side of the equality converges whenever $p>1$, by the integral test, the series converges for $p>1$.