Answer:
1. In order to make the integral improper
must be 1.
2. In order to make the integral improper
must be 1.
Explanation:
Using the rules of integration we get that for
![f(x)= (1)/(x^p)](https://img.qammunity.org/2021/formulas/mathematics/college/knww65wfljtuijijoubosbnqjjil6cucyc.png)
![\int\limits_(0)^(1) (1)/(x^p) \, dx = (x^(1-p))/((1-p)) \, |\limits_(0)^(1) = (1)/(1-p) - 0 = (1)/(1-p)](https://img.qammunity.org/2021/formulas/mathematics/college/zyuxmhkdmw0fdcr5x6ch3r58kmdkqs3leb.png)
Therefore in order to make that integral improper
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
![f(x) = (1)/(x(ln(x)^p))](https://img.qammunity.org/2021/formulas/mathematics/college/ak8jj6o2gpxrj9isq6e39fvoen9ztvsoxm.png)
![\int\limits_(e)^(\infty) (1)/(x(ln(x))^p) \, dx = ((ln(x))^(1-p))/(1-p) \, |\limits_(e)^(\infty)](https://img.qammunity.org/2021/formulas/mathematics/college/osuw6kjgicbhyytjziyltirrc93zm01kgv.png)
For
we would have
![\int\limits_(e)^(\infty) (1)/(x(ln(x))^p) \, dx = (1)/(p-1)](https://img.qammunity.org/2021/formulas/mathematics/college/4f4zxgc4xuxn4vfd3z2kefbbvgltv4l5eg.png)
And the problem is the same. If
we would have a 1/0 indeterminate form.