# $\int_{1}^{\infty} \dfrac{1}{x\ln(x + 1)} \ dx$ Comparison Test?

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.

• $\int_{1}^{N} \dfrac{1}{x\ln(x + 1)} \ dx\,>\int_{1}^{N} \dfrac{1}{(x+1)\ln(x + 1)} \ dx=\ln(\ln N)-\ln(\ln2)\, \to\infty$ as $N\to\infty$ Mar 28 at 20:59

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.

• Thank you very much!! I didn't think about what if you choose a factor of $k$ let's say. Mar 28 at 21:03

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.

• We cannot make a comparison in this way to show divergence because the inequality is in the wrong direction. Mar 28 at 20:52
• @heropup I've edited my answer. Thank you. Mar 28 at 21:01

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