How to show $\ \int_1^\infty\frac1xdx\ $ diverges (not using the harmonic series)?

Question

I was reading up on the harmonic series, $H=\sum\limits_{n=1}^\infty\frac1n$, on Wikipedia, and it's divergent, as can be shown by a comparison test using the fact that

$\begin{aligned}H&=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)+\cdots\\&\geq 1+\frac12+\left(\frac14+\frac14\right)+\left(\frac18+\frac18+\frac18+\frac18\right)+\cdots\\&=1+\frac12+\frac12+\frac12+\cdots,\end{aligned}$

where the expression on the right clearly diverges.

But after this proof idea was given, the proof idea using the integral test was given. I understand why $H_n=\sum_{k=1}^n\frac1k\geq \int_1^n \frac{dx}x$, but how is it shown that $\int_1^\infty \frac{dx}x$ is divergent without using the harmonic series in the following way: $H_n-1\leq \int_1^n \frac{dx}x\leq H_n$, and then using this in the following way, by comparison test:

$\lim\limits_{n\to\infty}H_n=\infty\implies\lim\limits_{n\to\infty}(H_n-1)=\infty\implies\lim\limits_{n\to\infty}\int_1^n \frac{dx}x=\infty$.

So to summarize, is there a way to prove that $\int_1^\infty \frac{dx}x$ without using the fact that $H$ diverges?

Answer 1

Let $x = y/2.$ Then

$$\int_1^\infty\frac{dx}{x} = \int_2^\infty\frac{dy}{y}.$$

That is a contradiction unless both integrals equal $\infty.$

Answer 2

Maybe this one. Change variables $x=t^{1/2}, dx = (1/2)t^{-1/2}\,dt$.

Then $$ \int_1^\infty \frac{1}{x}\;dx = \int_1^\infty \frac{1}{t^{1/2}}\;\frac{t^{-1/2}}{2}\;dt = \frac{1}{2}\int_1^\infty \frac{1}{t}\;dt $$ Now $\int_1^\infty \frac{dx}{x} > \int_1^2 \frac{dx}{2} = \frac{1}{2} > 0$. So conclude it is $+\infty$.

Or, if we are allowed properties of $e^x$:

Substitute $x=e^t, dx=e^t\,dt$ so $$ \int_1^\infty \frac{1}{x}\;dx = \int_0^\infty \frac{1}{e^t}\;e^t\;dt =\int_0^\infty dt = \infty. $$

Answer 3

Since $$ \int_a^{2a}\frac{\mathrm{d}x}x=\log(2) $$ we have that $$ \begin{align} \int_1^{2^n}\frac{\mathrm{d}x}x &=\int_1^2\frac{\mathrm{d}x}x+\int_2^4\frac{\mathrm{d}x}x+\cdots+\int_{2^{n-1}}^{2^n}\frac{\mathrm{d}x}x\\[6pt] &=n\log(2) \end{align} $$ Let $n\to\infty$.

Answer 4

$$\int_{ac}^{bc}\frac{dx}{x}=\int_{cx=ac}^{cx=bc}\frac{d(cx)}{cx}=\int_{a}^{b}\frac{dx}{x}$$ $$\therefore\int_{1}^{\infty}\frac{dx}{x}=\sum_{n=0}^\infty\int_{2^n}^{2^{n+1}}\frac{dx}{x}=\\ \sum_{n=0}^\infty\int_{1}^{2}\frac{dx}{x}=+\infty$$

Answer 5

This was pointed out in the comments above, so since no one else wrote this in an answer, I will.

You have (by definition) $$\begin{align} \int_1^\infty \frac{1}{x}\; dx &= \lim_{t\to \infty}\int_1^t \frac{1}{x}\; dx \\ &= \lim_{t\to \infty} \ln(x)\large]_1^t \\ &= \lim_{t\to \infty} \ln(t) - \ln(1) \\ &= \lim_{t\to \infty} \ln(t) \\ &= \infty. \end{align} $$

Answer 6

Note that

$$\begin{align} \int_1^{2^n}\frac{1}{u}\,du&=\int_1^2\frac{1}{u}\,du+\int_2^4\frac{1}{u}\,du+\int_4^8\frac{1}{u}\,du+\cdots +\int_{2^{n-1}}^{2^n}\frac{1}{u}\,du\\\\ &\ge \left(\frac{1}{1}\right)\left(2-1\right)+\left(\frac{1}{2}\right)\left(4-2\right)+\left(\frac{1}{4}\right)\left(8-4\right)+\cdots +\left(\frac{1}{2^{n-1}}\right)\left(2^n-2^{n-1}\right)\\\\ &=1+1+1+\cdots +1\\\\ &=n \end{align}$$

Therefore, we find that

$$\int_1^{2^n}\frac{1}{u}\,du\ge n \to \infty \,\,\text{as}\,\,n\to \infty$$

And we are done!