integrate from 5 to infinity of (ln(2x))/(x)dx Is the integral is divergent or convergent? If it is convergent, how do I evaluate it? If not, how do I find the answer?  

Expert Answers

An illustration of the letter 'A' in a speech bubbles

You need to check if the improper integral converges or diverges, hence, you need to evaluate if the following limit exists, such that:

`lim_(n->oo) int_5^n (ln(2x))/x dx  `

You need to use logarithmic identity such that:

`ln(a*b) = ln a + ln b`

Reasoning by analogy yields:

`ln(2x) = ln 2 + ln...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

You need to check if the improper integral converges or diverges, hence, you need to evaluate if the following limit exists, such that:

`lim_(n->oo) int_5^n (ln(2x))/x dx  `

You need to use logarithmic identity such that:

`ln(a*b) = ln a + ln b`

Reasoning by analogy yields:

`ln(2x) = ln 2 + ln x`

You need to divide by x such that:

`(ln(2x))/x = (ln 2)/x + (ln x)/x`

Integrating both sides yields:

`int (ln(2x))/x dx = int (ln 2)/x dx+ int (ln x)/x dx`

`int (ln(2x))/x dx = (ln 2) int 1/x dx + int (ln x)/x dx`

`int (ln(2x))/x dx = (ln 2)ln |x|+ int (ln x)/x dx`

You may solve the integral `int (ln x)/x dx`  using the following substitution such that:

`ln x = t => (1/x)dx = dt`

`int (ln x)/x dx = int t dt = t^2/2 + c`

Substituting back ln x for t yields:

`int (ln x)/x dx = (ln^2 x)/2 + c`

`int (ln(2x))/x dx = (ln 2) ln |x| + (ln^2 x)/2 + c`

You may evaluate the definite integral using the fundamental theorem of calculus such that:

`int_5^n (ln(2x))/x dx = ((ln 2) ln |x| + (ln^2 x)/2)|_5^n`

`int_5^n (ln(2x))/x dx = (ln 2)(ln n)+ (ln^2 n)/2 - (ln 2)(ln 5) - (ln^2 5)/2`

You need to evaluate the limit such that:

`lim_(n->oo) ((ln 2)(ln n) + (ln^2 n)/2 - (ln 2)(ln 5) - (ln^2 5)/2) = oo + oo - (ln 2)(ln 5) - (ln^2 5)/2 = oo`

Hence, evaluating the limit yields `lim_(n->oo) int_5^n (ln(2x))/x dx = oo` , hence, the improper integral diverges.

Approved by eNotes Editorial Team