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