Why does $\int_0^{\infty} \frac{1}{x}dx$ diverge?

Question

So I've only seen the p-test for integrals:

$\int_0^1$ and $\int_1^{\infty}$

here: http://www.sosmath.com/calculus/improper/testconv/testconv.html

but an example says that $\int_0^{\infty} \frac{1}{x}dx$ is also divergent.

I would like to know, how to know this.

Answer 1

For $0<a<b<\infty$, $\int_a^b\frac{1}{x}dx=\mathrm{log}b-\mathrm{log}a$. For $b\rightarrow \infty$ and $a\rightarrow 0$, it diverges to infinity.

Answer 2

Semi-formally: \begin{align*} \int_0^{+\infty} \frac{1}{x} \, dx &= \lim_{B \to +\infty} \lim_{A \to 0^+} \int_A^B \frac{1}{x} \, dx\\[0.3cm] &= \lim_{B \to +\infty} \lim_{A \to 0^+} \ln |x|\\[0.3cm] &= \lim_{B \to +\infty} \lim_{A \to 0^+} \left(\ln B - \ln A\right)\\[0.3cm] &= \lim_{B \to +\infty} \ln B - \lim_{A \to 0^+} \ln A\\[0.3cm] &= +\infty - (-\infty)\\[0.3cm] &= +\infty \end{align*}

Answer 3

Among the power functions $x\to x^\alpha$ ($x\in \mathbb{R^+}$), the case $\alpha=-1$ is an interesting limiting case.

For $\alpha>-1$, $\int_0^1 x^\alpha dx $ converges (and diverges for $\alpha<-1$). Morally, you gain one degree in the primitive, so somehow $\alpha+1>0$, and the primitive with shape $x^{\alpha+1}$ behaves ok at $0$.

Conversely, when $\alpha>-1$, $x^{\alpha+1}$ tends to infinity when $x\to \infty$, so $\int_1^{+\infty} x^\alpha dx $ diverges.

So with power functions, you are either good at $0$ or $+\infty$, never at both. Except for $\alpha=-1$ where you get the worse of both worlds, too fast blow-up at $0$, too slow decay at $+\infty$. The (c)onvergence or (d)ivergence with respect to $\alpha $ is summarized in this table: \begin{array} {|r|c|c|c|} \hline \alpha& < -1& = -1 & >-1\\ \hline \int_0^1 x^{\alpha} dx& \text{d}& \text{d}& \text{c}\\ \hline \int_1^{+\infty}x^{\alpha} dx & \text{c}& \text{d}& \text{d}\\ \hline \end{array}

One of it primitive is $\log x$, and to some extend, the logarithm is a twisted sort of unitless $0$th power function.

Answer 4

Another way to look at this is to notice that $$ \int_0^\infty \frac{1}{x} dx \ge \int_0^{\infty} \sum_{N=1}^M f_N(x) dx, $$

where $ f_N(x) = 1/N $ if $ (N-1) \le x \le N $ and $ f_N(x) = 0 $ otherwise. The integral $ \int_0^\infty f_N(x) dx = 1/N $ for each $ N \ge 1 $, so once you justify interchanging sum and integral (simply just by linearity of the integral), you see that

$$ \lim_{M \to \infty} \int_0^\infty \sum_{N=1}^M f_N(x) dx = \sum_{N=1}^\infty \frac{1}{N}. $$

This infinite sum is divergent because it is the harmonic series, so we see $$\int_0^\infty \frac{1}{x} dx \ge \sum_{N=1}^\infty \frac{1}{N} = \infty. $$

Answer 5

If $\int_1^\infty \frac 1x dx$ were convergent, then by doing a change of variable $x=2t$, you would have
$$\int_1^\infty \frac 1x dx = \int_{1/2}^\infty \frac 1{2t} 2dt = \int_{1/2}^\infty \frac 1t dt = \int_{1/2}^1 \frac 1t dt + \int_1^\infty \frac 1x dx$$.

And so $\int_{1/2}^1 \frac 1t dt = 0$. But this is clearly impossible since $\int_{1/2}^1 \frac 1t dt \ge \int_{1/2}^1 1 dt = 1/2 > 0$