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

$H=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+...\geq 1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+...=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...,$ 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 $H_n=\sum_{k=1}^n \frac{1}{k}\geq \int_1^n \frac{dx}{x}$, 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:

$\lim_{n\rightarrow\infty}H_n=\infty\Rightarrow\lim_{n\rightarrow\infty}(H_n-1)=\infty\Rightarrow\lim_{n\rightarrow\infty}\int_1^n \frac{dx}{x}=\infty$.

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

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    $\begingroup$ A primitive of $\frac{1}{x}$ is $\log x$, so $\int_a^b \frac{dx}{x} = \log b - \log a$ for $0 < a < b$. $\endgroup$ – Daniel Fischer Jul 16 '15 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 '15 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. $\endgroup$ – Khadija Mbarki Jul 16 '15 at 23:11
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    $\begingroup$ 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 '15 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 $\log x > K$]. $\endgroup$ – Daniel Fischer Jul 16 '15 at 23:13
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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 $\infty.$

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  • $\begingroup$ Nice indeed. +1 $\endgroup$ – Timbuc Jul 16 '15 at 23:17
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    $\begingroup$ +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 '15 at 23:28
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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 $n\to\infty$.

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

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  • $\begingroup$ 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 '18 at 13:25
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This was pointed out in the comments above, so since no one else wrote this in an answer, I will.

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

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  • $\begingroup$ This seems to be the simplest way to do it! I'm surprised no one posted it before you. $\endgroup$ – Skeleton Bow Oct 2 '16 at 19:07
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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!

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

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