Answer:
Step-by-step explanation:
Consider the integral
[tex]\int\limits^1_{-infty} \frac{1}{x^p} \, dx[/tex]
for various values of p
CaseI:p =1 Then integral is ln x and for infinity this value diverges so diverges
Case 2: p<1
The integrated value would be [tex]\frac{x^{-p+1} }{-p+1}[/tex]
Since P <1 numerator x will have positive exponent and when infinity is substituted this will diverge
Case 3: p >1
[tex]\frac{x^{-p+1} }{-p+1}[/tex] will have x with negative exponent. So when x = infinity this value becomes 0 thus making the integral to converge
