Find the value of p, improper integral

**racewithferrari** · February 25th 2010, 05:59 PM

Find the value of p for which the integral converges and evaluate the integral for those values of p

∫ 1/(x(ln x)^p)

upper limit : infinity
lower limit : e

**Scott H** · February 25th 2010, 06:05 PM

Our integral is

$\int_0^{\infty}\frac{1}{x(\ln x)^p}\,dx=\int_0^{\infty}\frac{1}{x}\cdot\frac{1}{ (\ln x)^p}\,dx.$

Hint: we may use a substitution of variables.

**racewithferrari** · February 25th 2010, 06:17 PM

I didn't really get it. can you explain it further. BTW there was a typo, the lower limit is e not 0.

**Krizalid** · February 25th 2010, 08:03 PM

well that really makes a difference, so make the substitution what do you get?

what can we say about $\int_1^\infty\frac{dt}{t^p}$ ?

**chisigma** · February 25th 2010, 09:22 PM

The integral proposed by Scott H , though different from the original one, is quite interesting. First we suppose that is $p\ne 1$ . In this case the integral can be devided in two parts...

$\int_{0}^{\infty} \frac{dx}{x\cdot \ln^{p} x}= \int_{0}^{1} \frac{dx}{x\cdot \ln^{p} x} + \int_{1}^{\infty} \frac{dx}{x\cdot \ln^{p} x}$ (1)

Now we can set in second integral $t=\frac{1}{x}$ and obtain...

$\int_{1}^{\infty} \frac{dx}{x\cdot \ln^{p} x} = (-1)^{p} \int_{0}^{1} \frac{dt}{t\cdot \ln^{p} t}$ (2)

... so that is...

$\int_{0}^{\infty} \frac{dx}{x\cdot \ln^{p} x}= \{1 + (-1)^{p}\}\cdot \int_{0}^{1} \frac{dx}{x\cdot \ln^{p} x} = \lim_{\xi \rightarrow 1} \frac{\{1 + (-1)^{p}\}}{p-1} \cdot \ln ^{1-p} \xi$ (3)

From (3) we deduct that for $p$ odd with $p\ne 1$ is...

$\int_{0}^{\infty} \frac{dx}{x\cdot \ln^{p} x}= 0$ (4)

... and for $p$ even with $p \ge 2$ the integral diverges...

Kind regards

$\chi$ $\sigma$

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