# The integrals from $1$ to $\infty$ for $\dfrac{1}{x}$ and $\dfrac{1}{x^2}$

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

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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". – kahen Nov 1 '12 at 20:14
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. – Hurkyl Nov 1 '12 at 20:20

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. – Jarvix Nov 1 '12 at 20:19

\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 $\epsilon \to 0$. – gt6989b Nov 1 '12 at 20:30

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) – Hurkyl Nov 1 '12 at 20:19
@Hurkyl Yes, hence the link. I did not want to write it all out. – glebovg Nov 1 '12 at 20:21
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 – Jarvix Nov 1 '12 at 20:21
@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. – Hurkyl Nov 1 '12 at 20:22
@Hurkyl Thanks. – glebovg Nov 1 '12 at 20:24

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.

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