Is the area under this graph finite?

Question

Given the graph of $y=\frac{1}{x}$, and assuming that we limit our graph to the first quadrant, i.e. $x≥0$ and $y≥0$

Is the area under the curve finite? If so, what is the maximum?

Answer 1

No, the area under this graph is infinite, as shown here:

$\int\limits_0^\infty \frac{1}{x}\geq \sum\limits_{n=1}^\infty \frac{1}{n} = \lim\limits_{n\to\infty}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}\right)$

$\geq \lim\limits_{n\to\infty}\left(\frac{1}{1}+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+(\frac{1}{16}+\dots+\frac{1}{16})+\dots+(\frac{1}{2^n}+\dots+\frac{1}{2^n})\right)$

$=\lim\limits_{n\to\infty}1+\underbrace{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots+\frac{1}{2}}_{n~\text{times}}=\infty$

Answer 2

No, this is incorrect. Were the area finite, $\log(x) = \int_{1}^{x} \frac{dx}{x}$ would be bounded. But this is absurd, since $\exp$ has domain $\mathbb{R}$.

Answer 3

$\int_0^\infty \frac{1}{x}dx=[ln(x)]_0^\infty=ln(\infty)-ln(0)=\infty+\infty=\infty$

Of course $ln(0)$ doesn't exist, but $\lim\limits_{x \to 0+} ln(x)=-\infty$

Therefore the surface under $y=\frac{1}{x}$ is infinite.

Answer 4

No because the series: $$\sum\limits_{i=1}^{\infty} \frac{1}{i} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + .... $$ Is a diverging series, specifically the harmonic series. There are multiple proofs out there for this.

Answer 5

For each positive integer $n$, consider the rectangle in the Cartesian coordinate plane with vertices $$\{(n-1, 0), (n, 0), (n, 1/n), (n-1, 1/n)\};$$ that is to say, define regions $$R_n = \{ (x,y) \in \mathbb R^2 : (n-1 < x \le n) \cap (0 \le y \le 1/n)\}.$$ Each such rectangle has the vertex $(n, 1/n)$ which obviously lies on the curve $y = f(x) = 1/x$. Furthermore, because $1/(n-1) > 1/n$, we also see that each rectangle is contained within the region $$R = \{(x,y) \in \mathbb R^2 : 0 \le xy \le 1 \}.$$ Now the area of $R_n$ is $$A_n = |R_n| = (n - (n-1)) \times (1/n - 0) = 1/n,$$ and it is also clearly true that $R_i \cap R_j = \emptyset$ whenever $i \ne j$. Therefore, $$|R| > \sum_{n=1}^\infty A_n = \sum_{n=1}^\infty \frac{1}{n}.$$ Now it is straightforward to see that $$H_{2^m} = \sum_{n=1}^{2^m} \frac{1}{n} = \sum_{n=1}^{2^{m-1}} \frac{1}{n} + \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n} > H_{2^{m-1}} + \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{2^m} = H_{2^{m-1}} + \frac{2^{m-1}}{2^m} = \frac{1}{2} + H_{2^{m-1}},$$ consequently $H_{2^m} > \frac{m}{2} + 1$ for all positive integers $m$, and there is no positive integer $N$ for which $H_{2^m} < N$ for all $m$. Hence $|R|$ is not finite.