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So I am having a bit of trouble with this integral, and I would like some verification to see if I am on the right track.

Q: Determine if $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ converges or diverges.

So I know that we need to find a function that is comparable to the original function. So I chose $\displaystyle \dfrac{1}{x\ln(x)}$ because \begin{equation*} \dfrac{1}{x\ln(x + 1)} \leq \dfrac{1}{x\ln(x)} \end{equation*} for $x \geq 1$.

So we take the integral: $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x)} \ dx$. Let $u = \ln(x)$, then $du = \dfrac{1}{x} \ dx$. \begin{align*} \int_{1}^{\infty} \dfrac{1}{x\ln(x)} \ dx &= \int_{x = 1}^{x = \infty} \dfrac{1}{u} \ du \\ &=\lim_{t \to \infty}\left[\ln|u|\right]_{x = 1}^{x = t} \\ &= \lim_{t \to \infty} \left[\ln|\ln(x)|\right]_{1}^{t} \\ &= \lim_{t \to \infty} \ln|\ln(t)| - \lim_{t \to \infty} \ln|\ln(1)| \\ &= \infty - DNE \end{align*} The last part is where I am having a bit of trouble. I know that the integral diverges with the infinity, which means that $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ also diverges. But if $\ln|\ln(1)|$ does not exist, does that mean this integral still diverges? I am not sure how to explain this part well. Would appreciate some tips.

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  • $\begingroup$ $\int_{1}^{N} \dfrac{1}{x\ln(x + 1)} \ dx\,>\int_{1}^{N} \dfrac{1}{(x+1)\ln(x + 1)} \ dx=\ln(\ln N)-\ln(\ln2)\, \to\infty$ as $ N\to\infty$ $\endgroup$ – Svyatoslav Mar 28 at 20:59
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If you suspect the integral diverges, you must compare it with another integral that is smaller and show that the smaller one diverges. Otherwise, you could compare for instance $$\int_{x=1}^\infty \frac{1}{x^2} \, dx < \int_{x=1}^\infty \frac{1}{x} \, dx,$$ and the RHS is divergent, but that tells you nothing about whether the LHS is convergent or divergent.

As such, we would need to compare $$f(x) = \frac{1}{x \log(x+1)}$$ against some other function, say $g(x)$, that is uniformly smaller, but its integral on the same interval is divergent. Your choice $$g(x) = \frac{1}{x \log x}$$ does not work. Instead, I would choose

$$g(x) = \frac{1}{x \log 2x}.$$ Then on $[1, \infty)$, we easily see $$\frac{1}{x \log 2x} \le \frac{1}{x \log (x+1)},$$ hence $$\int_{x=1}^\infty f(x) \, dx \ge \int_{x=1}^\infty g(x) \, dx.$$ I leave it as an exercise to evaluate the antiderivative of $g$ and show the definite integral is unbounded.

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  • $\begingroup$ Thank you very much!! I didn't think about what if you choose a factor of $k$ let's say. $\endgroup$ – user858034 Mar 28 at 21:03
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That integral converges if and only if the integral$$\int_2^\infty\frac1{x\log(x+1)}\,\mathrm dx.$$But we have$$\lim_{x\to\infty}\frac{\frac1{x\log(x+1)}}{\frac1{x\log(x)}}=1\tag1$$But the integral$$\int_2^\infty\frac1{x\log(x)}\,\mathrm dx=\lim_{M\to\infty}\log(\log(M)))-\log(\log(2))=\infty.$$So, $\int_2^\infty\frac1{x\log(x)}\,\mathrm dx$ diverges, and therefore it follows from $(1)$ that your integral diverges too.

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  • $\begingroup$ We cannot make a comparison in this way to show divergence because the inequality is in the wrong direction. $\endgroup$ – heropup Mar 28 at 20:52
  • $\begingroup$ @heropup I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Mar 28 at 21:01
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OP wrote: I know that the integral diverges with the infinity, which means that $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ also diverges.

You can't figure this out of comparison test. The comparison test says that: When for all $x$ we have $0 ≤ f(x) ≤ g(x)$, if $\displaystyle \int f(x) \ dx$ diverges, than $\displaystyle \int g(x) \ dx$ diverges, but not the opposite!

I think this is what you're looking for $\begin{equation*} \dfrac{1}{(x+1)\ln(x + 1)} \leq \dfrac{1}{x\ln(x+1)} \end{equation*}$.

After that you can simply substitute $x+1=t$ and figure out that $\displaystyle \int_{1}^{\infty} \dfrac{1}{(x+1)\ln(x+1)} \ dx$ diverges!

Now it's simple to figure out that $\displaystyle \int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ diverges from the comparison test

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