proof of integral diverging

Question

So the idea behind this proof is to use limits of definite integrals.

Question: If

0$\lt$b $\leqslant$1, then $$\int_1^{\infty}\frac{1}{x^b}dx$$ diverges.

I was wondering how might this proof be done? What does it mean by using th elimits of the definite integral.

Answer 1

HINT

Note that by definition

$$\int_1^\infty {1\over x}dx = \lim_{a\to\infty} \int_1^a {1\over x}dx$$

and

• $0<b<1 \implies\int \frac{1}{x^b}dx=\frac{x^{1-b}}{1-b}+c$
• $b=1 \implies\int \frac{1}{x}dx=\ln x+c$

Answer 2

$$\lim_{M\to +\infty} \int_1^M \frac{1}{x^b}\ dx = \frac{x^{1-b}}{1-b}\bigg|_1^M = \frac{M^{1-b}}{1-b}$$

You can easily see that as $M\to +\infty$ the integral does diverge, unless $1-b < 0$ that is, $b>1$ (which is not your case).