The integrals from $1$ to $\infty$ for $\frac{1}{x}$ and $\frac{1}{x^2}$ I have two integrals:
$$
A=\int\limits_1^\infty \dfrac{1}{x}dx\,,
B=\int\limits_1^\infty \dfrac{1}{x^2}dx
$$
Calculus says that A is an improper integral as it diverges, but the B converges and is $1$, because $\dfrac{1}{x^2}$ is faster near $y=0$ than $\dfrac{1}{x}$.
I don't understand the reason behind this. So I looked for another way to put down my problem. Multiple sources define that:
$$
\frac{1}{\infty} = \frac{1}{\infty^2}
$$
What is the reason that the integral of $\dfrac{1}{x}$ is divergent and $\dfrac{1}{x^2}$ is convergent? In the end they both reach $\dfrac{1}{\infty}$ (or $\dfrac{1}{\infty^2}$ which is $\dfrac{1}{\infty}$).
 A: The answer has nothing to do with '$\infty$ calculus'. Calculate the finite integral first, and take limits.
$\int_1^x \frac{1}{t} dt = \ln x$, $\int_1^x \frac{1}{t^2} dt = 1-\frac{1}{x}$.
$A = \lim_{x \to \infty} \ln x = \infty$, $B =\lim_{x \to \infty} 1-\frac{1}{x} = 1$.
A: \begin{eqnarray}
\int\limits_1^\infty \dfrac{1}{x}dx & = & \lim_{ \varepsilon \rightarrow \infty }   \int\limits_1^\varepsilon \dfrac{1}{x}dx 
& = & \lim_{ \varepsilon \rightarrow \infty } \log \varepsilon = \infty
\end{eqnarray}
While
\begin{eqnarray}
\int\limits_1^\infty \dfrac{1}{x^2}dx & =& \lim_{ \varepsilon \rightarrow \infty } \Bigr(-\dfrac{1}{\varepsilon } +1 \Bigl) = 1
\end{eqnarray}
A: There is something called Cauchy integral test which tells us that we can compare an infinite sum with a corresponding integral and vice versa under certain circumstances (see the link). Because
$$\sum\limits_{n = 1}^\infty  {\frac{1}{x}}$$
diverges so does the integral and since
$$\sum\limits_{n = 1}^\infty  {\frac{1}{{{x^2}}}}$$
converges so does the integral. By the way, infinity is not a number. Saying $\infty^2$ is the same as saying $\text{blue}^2$ -- it does not mean anything, so you cannot look at these integral as fractions with $\infty$ as a denominator.
A: Improper integrals, such as $A$ and $B$ are defined as limits:
$$A= \int_1^{\infty} \frac{1}{x}dx= \lim_{R \to \infty} \int_1^R \frac{1}{x}dx$$
and 
$$B= \int_1^{\infty} \frac{1}{x^2}dx= \lim_{R \to \infty} \int_1^R \frac{1}{x^2}dx$$
If you carry out the details you'll get $$\lim_{R \to \infty} \ln R$$ which goes to $\infty$ and $$\lim_{R \to \infty} \frac{1}{R}$$ which converges.
