How do you use the integral test to determine whether int dx/lnxdxlnx converges or diverges from [2,oo)[2,)?

1 Answer
Jun 25, 2018

The integral

int_2^oo dx/lnx2dxlnx

is divergent.

Explanation:

Note that in the interval x in[2,oo)x[2,) the function:

f(x) = 1/lnxf(x)=1lnx

is:

1) Infinitesimal as lim_(x->oo) f(x) = 0

2) Positive as f(x) >0 for x >1

3) Decreasing. In fact f'(x) = -1/(xln^2x) < 0

4) f(n) = 1/lnn

So, based on the integral test, the convergence of the integral:

int_2^oo dx/lnx

is equivalent to the convergence of the series:

sum_(n=2)^oo 1/lnn

Now we can easily demonstrate that:

lnn < n

so that:

1/ln n > 1/n

and as we know that the harmonic series:

sum_(n=1)^oo 1/n

is divergent, we can conclude that:

sum_(n=2)^oo 1/lnn

is divergent by direct comparison, and hence also:

int_2^oo dx/lnx

is divergent.