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?
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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.
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