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
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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.