Show that the converges $$\int_{∞}^{1}\frac{1}{x^2} \,dx$$ and diverges $$\int_{∞}^{1}\frac{1}{x}\,dx$$

Solution:

If limit of  $$\int_{∞}^{1}\frac{1}{x^2} \,dx$$ exists then it converges

= $$\int_{∞}^{1}\frac{1}{x^2} \,dx$$

= $$\int_{∞}^{1} x^{-2} \,dx$$

= $$\begin{bmatrix}\frac{x^{-1}}{-1}\end{bmatrix}^{\!1}_{∞}$$

= $$\begin{bmatrix}\frac{-1}{x}\end{bmatrix}^{\!1}_{∞}$$

= -1 – $$\frac{1}{∞}$$

= -1

Since the limit exists so it converges.

Second Part:

If limit of  $$\int_{∞}^{1}\frac{1}{x} \,dx$$ doesn’t exists then it diverges

= $$\int_{∞}^{1}\frac{1}{x} \,dx$$

= $$\begin{bmatrix} log(x) \end{bmatrix}^{\!1}_{∞}$$

= log(1) – log(∞)

= 0 – ∞

= – ∞

Since the limit doesn’t exists so it diverges.