Answer :
Answer:
1. In order to make the integral improper [tex]p[/tex] must be 1.
2. In order to make the integral improper [tex]p[/tex] must be 1.
Step-by-step explanation:
Using the rules of integration we get that for
[tex]f(x)= \frac{1}{x^p}[/tex]
[tex]\int\limits_{0}^{1} \frac{1}{x^p} \, dx = \frac{x^{1-p}}{(1-p)} \, |\limits_{0}^{1} = \frac{1}{1-p} - 0 = \frac{1}{1-p}[/tex]
Therefore in order to make that integral improper [tex]p[/tex] must be 1.
If p = 1 then you would have a 1/0 indeterminate form.
2. Using the of integration, specifically substitution we get that for
[tex]f(x) = \frac{1}{x(ln(x)^p)}[/tex]
[tex]\int\limits_{e}^{\infty} \frac{1}{x(ln(x))^p} \, dx = \frac{(ln(x))^{1-p}}{1-p} \, |\limits_{e}^{\infty}[/tex]
For [tex]p \geq 1[/tex] we would have
[tex]\int\limits_{e}^{\infty} \frac{1}{x(ln(x))^p} \, dx = \frac{1}{p-1}[/tex]
And the problem is the same. If [tex]p=1[/tex] we would have a 1/0 indeterminate form.