Question

Determine which values of p the following integrals converge. Give your answer in each case by selecting the appropriate inequality and then entering a numerical value in the box to define a range of p values for which the integral converges. If the integral never converges, select = and enter none for the numerical value. In either case be sure that you can rigorously prove convergence and divergence for different values of p.

a) integral from 1 to 2 of (dx/(x(ln(x))^p)

b) integral from 0 to 1 of (ln(x)dx/x^p)

Answer 1

Answer:

Step-by-step explanation:

a)

$\int\limits^2_1 {\frac{1}{x(lnx)^p} } \, dx$

this can be done by substitute lnx = u

dx/x = du

When x =1, u =0 and when x =2, u = ln 2

So integral = $\int\limits^{ln2} _0 {du/u^p} \\\=\frac{u^{-p+1} }{-p+1}$

We find that this integral value is not definid for p =1

Hence for values of p other than 1, this converges.

When we substitute limits

$\frac{1}{1-p} ((ln2)^{1-p} -1)$

and converges for p ≠1

b) $\int\limits^1_0 {lnx}/x^p \, dx \\\int \frac{\ln \left(x\right)}{x^p}dx=\frac{1}{-p+1}x^{-p+1}\ln \left(x\right)-\frac{x^{-p+1}}{\left(-p+1\right)^2}+C$

So not converging for p =1

But ln x is defined only for x >0

So integral 0 to 1 makes this integral not valid and hence not convergent.