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.
Best Answer
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.