# Thread: Find the value of p, improper integral

1. ## Find the value of p, improper integral

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

2. Our integral is

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

Hint: we may use a substitution of variables.

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

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

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

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

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

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

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

... so that is...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$