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

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?

• A primitive of $\frac{1}{x}$ is $\log x$, so $\int_a^b \frac{dx}{x} = \log b - \log a$ for $0 < a < b$. Commented Jul 16, 2015 at 23:06
• 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? Commented Jul 16, 2015 at 23:09
• 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. Commented Jul 16, 2015 at 23:11
• 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!
– anon
Commented Jul 16, 2015 at 23:12
• 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 $\log x > K$]. Commented Jul 16, 2015 at 23:13

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 $\infty.$

• Nice indeed. +1 Commented Jul 16, 2015 at 23:17
• +1 very nice! Of course it works because $\int_1^2 \frac 1x dx \neq 0$ which is obvious but worth mentioning I think :)
– Ant
Commented Jul 16, 2015 at 23:28
• 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 Commented 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 $$\int_1^\infty \frac{1}{x}\;dx = \int_0^\infty \frac{1}{e^t}\;e^t\;dt =\int_0^\infty dt = \infty.$$

• 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 $0$.
– robjohn
Commented 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}

• This seems to be the simplest way to do it! I'm surprised no one posted it before you. Commented 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!

• 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 Commented Mar 14, 2023 at 9:14

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