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?

  • 11
    $\begingroup$ A primitive of $\frac{1}{x}$ is $\log x$, so $\int_a^b \frac{dx}{x} = \log b - \log a$ for $0 < a < b$. $\endgroup$ Jul 16, 2015 at 23:06
  • $\begingroup$ Yes, but then there is the question of how to prove that $\log x\rightarrow\infty$ as $x\rightarrow \infty$. Is it possible to do without using the harmonic series, like somehow using the fact that the logarithm is the inverse of the exponential function? $\endgroup$
    – Scounged
    Jul 16, 2015 at 23:09
  • $\begingroup$ Maybe you can use the Bertrand criterion for the integral $\int_{1}^{\infty} \frac{1}{t^{\alpha} \log(t)^{\beta}} dt$ with $\alpha =1$ and $\beta =0$ so you get the divergence of the integral. $\endgroup$ Jul 16, 2015 at 23:11
  • 2
    $\begingroup$ Given $N$, we know $\log x>N$ for all $x>e^N$. So yes, $\log x$ diverges. Your logic concerning the harmonic series is backwards, BTW. We do not use the harmonic series' divergence to prove $\int_1^\infty\frac{1}{x}dx$ or $\log x$ diverges - we use the latter to prove the former! $\endgroup$
    – anon
    Jul 16, 2015 at 23:12
  • $\begingroup$ Yes, using that it's the inverse of $\exp$ and noting that $\lim\limits_{x\to +\infty} \exp(x) = +\infty$ shows that $\lim\limits_{x\to +\infty} \log x = +\infty$ [for $x > e^K$, we have $\log x > K$]. $\endgroup$ Jul 16, 2015 at 23:13

6 Answers 6


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 $\infty.$

  • 1
    $\begingroup$ Nice indeed. +1 $\endgroup$
    – Timbuc
    Jul 16, 2015 at 23:17
  • 17
    $\begingroup$ +1 very nice! Of course it works because $\int_1^2 \frac 1x dx \neq 0$ which is obvious but worth mentioning I think :) $\endgroup$
    – Ant
    Jul 16, 2015 at 23:28
  • $\begingroup$ This is a nice proof. But proving the substitution rule for improper integrals would probably take more work than another simple way of proving the integral diverges. Regardless, the proof is pretty interesting $\endgroup$ Mar 14, 2023 at 9:10

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 $n\to\infty$.


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

  • 1
    $\begingroup$ The first approach shows either that the integral is $0$ or that the integral diverges. It might be worth mentioning, if obvious, that the integral is not $0$. $\endgroup$
    – robjohn
    May 24, 2018 at 13:25

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

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

  • $\begingroup$ This seems to be the simplest way to do it! I'm surprised no one posted it before you. $\endgroup$ Oct 2, 2016 at 19:07

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!

  • $\begingroup$ Nice proof! Though you have got the inequalities wrong, each term is less than or equal to the function evaluated at the right end point (since it is a decreasing function) times the length of the interval, so you get the bound n/2. Doesn't change the idea just a mistake $\endgroup$ Mar 14, 2023 at 9:14

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .