Determine which values of p the following integral converge. Give your answer in each case by giving an appropriate inequality sign (e.g., > or <=) to define a range of p values for which the...
Determine which values of p the following integral converge. Give your answer in each case by giving an appropriate inequality
sign (e.g., > or <=) to define a range of p values for which the integral converges. If the integral never converges, enter none for the numerical value.
integrate from 2 to infinity of ((dx)/(x(ln(x)))^p)
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You should solve the given improper integral such that:
`int_2^oo 1/(x*(ln x)^p)dx = lim_(n->oo) int_2^n 1/(x*(ln x)^p)dx`
You need to solve the indefinite integral `int 1/(x*(ln x)^p)dx` using the substitution `ln x = t` such that:
`ln x = t => 1/x dx = dt`
Changing the variable yields:
`int 1/(x*(ln x)^p)dx = int (dt)/(t^p)`
Using the property of negative power yields:
`1/t^p = t^(-p)`
`int (dt)/(t^p) = intt^(-p) dt =(t^(-p+1))/(-p+1) + c`
Using the fundamental theorem of calculus yields:
`int_2^n 1/(x*(ln x)^p)dx = ((ln x)^(-p+1))/(-p+1)|_2^n`
`int_2^n 1/(x*(ln x)^p)dx = ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p)`
Evaluating the limit yields:
`lim_(n->oo) ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p) = lim_(n->oo) ((ln n)^(1-p))/(1-p) - lim_(n->oo)((ln 2)^(1-p))/(1-p) `
If `p>1 => lim_(n->oo)((ln n)^(1-p))/(1-p) = lim_(n->oo) 1/(((ln n)^(p-1))(1-p)) = 1/oo = 0`
`lim_(n->oo) ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p) =0 - ((ln 2)^(1-p))/(1-p) = - ((ln 2)^(1-p))/(1-p)`
If `p<1 => lim_(n->oo) ((ln n)^(1-p))/(1-p) - ((ln 2)^(1-p))/(1-p) = oo`
Hence, evaluating the given improper integral yields that integral converges if p>1 and integral diverges that p<1.
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