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s2008m [1.1K]
3 years ago
7

Determine which values of p the following integrals converge. Give your answer in each case by selecting the appropriate inequal

ity 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)
Mathematics
1 answer:
sashaice [31]3 years ago
7 0

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.

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