Question

For what values of $p$, is the integral ${\int }_1^{\infty }\frac{dx}{x^p}$ is convergent?

• A. $p>1$
• B. $p>-1$
• C. $p<1$
• D. None of these

Answer 1

Answer: (A)

$p>1$

Given : ${\int }_1^{\infty }\frac{dx}{x^p}$

So to find its convergent value

${\int }_1^{\infty }\frac{dx}{x^p}=\stackrel{lim}{t\to \infty }{\int }_1^t{x}^{-p}dx={\left[\frac{x^{1-p}}{1-p}\right]}_1^t=\frac{t^{1-p}}{1-p}=\frac{1}{(1-p)}$

Sign of $1-p$ is positive when $p<1$

And the scene will then be divergent

Sign of $1-p$ is negative when $p>1$

And $\stackrel{lim}{t\to \infty }\frac{t^{1-p}}{1-p}=\frac{1}{(1-p)}=0$ Convergent for $(p>1)$

Hence the correct answer is $p>1$