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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 $\dfrac{1}{\infty}$).

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    They're *both* improper (Riemann) integrals. $A$ is a divergent improper Riemann integral and $B$ is a convergent one. The Riemann integral is only defined for (certain) bounded functions on bounded intervals. They're called improper because they're limits (in a suitable sense) as the (in this case) upper integration boundary "goes to infinity".2012-11-01
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    What happens *at* $\infty$ doesn't matter. It's what happens "near" $\infty$ that matters: i.e. how the function behaves for large (finite) real values.2012-11-01

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The answer has nothing to do with '$\infty$ calculus'. Calculate the finite integral first, and take limits.

$\int_1^x \frac{1}{t} dt = \ln x$, $\int_1^x \frac{1}{t^2} dt = 1-\frac{1}{x}$.

$A = \lim_{x \to \infty} \ln x = \infty$, $B =\lim_{x \to \infty} 1-\frac{1}{x} = 1$.

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    Thanks! This is exactly what I was looking for.2012-11-01
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\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}

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    You likely mean $\epsilon \to \infty$, not $\epsilon \to 0$.2012-11-01
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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.

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    (assuming the function $f$ is non-negative, decreasing, and continuous)2012-11-01
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    @Hurkyl Yes, hence the link. I did not want to write it all out.2012-11-01
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    I know it is a label, but as it is infinite, there is nothing more, so why would $\infty^2$ not be $\infty$? And is Wolfram|Alpha wrong? [link](http://www.wolframalpha.com/input/?i=1%2Finfinity+%3D+1%2Finfinity%5E2)2012-11-01
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    @glebovg: It's worth mentioning something conditional in your statement, then. e.g. "... which tells us **when** we can compare ..." or "... which tells us that we can compare ... if certain properties are met ..." or something.2012-11-01
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    @Hurkyl Thanks.2012-11-01
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    @Jarvix It is a word. $\infty$ is a concept. For example taking the limit as something approached infinity is not the same as evaluating it at infinity. Not to mention, [some infinities are bigger than others](http://mathworld.wolfram.com/CantorDiagonalMethod.html).2012-11-01
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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.