Conjecture regarding sequence of improper integrals of form $\int_1^\infty \frac{\ln x}{f(x)}dx$ I was playing around with integrals of the form $$\int_1^{\infty}\frac{\ln x}{f(x)}dx$$
and noticed something interesting when $f(x)=x(x-1)$. By numerical computation the answer tends to $1.644\ldots$ which is $\frac{\pi^2}{6}$. By complete chance, I tried setting $f(x)=x^2(x-1)$ which yielded $0.644\ldots$ which is $\frac{\pi^2}{6}-1$. I continued increasing the power of $x$, yielding $0.394\ldots (\frac{\pi^2}{6}-1-0.25)$, then $0.2838\ldots (\approx\frac{\pi^2}{6}-1-0.25-0.11)$. Define $$I_n=\int_1^{\infty}\frac{\ln x}{x^n(x-1)}dx.$$ From the pattern, my conjecture is $$I_n=\frac{\pi^2}{6}-\sum_{m=1}^{n-1}m^{-2}=\sum_{m=n}^{\infty}m^{-2}.$$
I'm trying to prove this, and have so far considered
$$
\begin{align}
I_n-I_{n+1}&= \int_1^{\infty}\frac{\ln x}{x^n(x-1)}dx-\int_1^{\infty}\frac{\ln x}{x^{n+1}(x-1)}dx
\\ &=\int_1^{\infty}\frac{\ln x}{x-1}\left(\frac{x-1}{x^{n+1}}\right)dx
\\ &= \int_1^{\infty}\frac{\ln x}{x^{n+1}}dx
\\ &= \lim_{k\to\infty}\left[\frac{-\ln x}{nx^n}\right]^{k}_{1}+n^{-1}\int_{1}^\infty x^{-n-1}dx
\\ &= -n^{-1}\lim_{k\to\infty}\left(\frac{\ln k}{k^n}\right)+n^{-1}\left[\frac{-1}{nx^n}\right]_1^\infty
\\&=-n^{-1}\underbrace{\lim_{k\to\infty}\left(\frac{k^{-1}}{nk^{n-1}}\right)}_{\text{By L'Hopital's rule}}+n^{-2}(0--1)
\\&= -n^{-1}\underbrace{\lim_{k\to\infty}\left(\frac{1}{nk^{n}}\right)}_{=0}+n^{-2}
\\ &= n^{-2}.
\end{align}
$$
Now we can take the partial sum to $N$ (over all $1\leqslant n \leqslant N$), and take the limit as $N\to\infty$, like so.
$$
\begin{align}
\lim_{N\to\infty}\sum_{n=1}^{N} I_n-I_{n+1}&=\lim_{N\to\infty}\left(\sum_{n=1}^{N} I_n-\sum_{n=2}^{N+1}I_{n}\right)
\\\iff \lim_{N\to\infty}\sum_{n=1}^{N}n^{-2} &=\lim_{N\to\infty}(I_1-I_{N+1})
\\\iff \sum_{n=1}^\infty n^{-2} &= I_1-\lim_{N\to\infty}\int_1^{\infty}\frac{\ln x}{x^{N+1}(x-1)}dx
\\\iff I_1&=\frac{\pi^2}{6}+\lim_{N\to\infty}\int_1^{\infty}\frac{\ln x}{x^{N+1}(x-1)}dx
\end{align}
$$
So my issue is I'm unsure how to go about evaluating the limit of the integral. The integrand clearly tends to 0 for any $x\in (1,\infty)$ when $N$ is arbitrarily large, which would mean the integral tends to 0 (which is what would lead to the completion of the proof, after an induction on $n$), but for some reason this feels very unrigorous (especially because the integral itself is improper).
Also does anyone have any different approaches to find $I_n$? Perhaps there's a solution using Taylor series? It seems like quite a "neat" formula (in the conjecture) for $I_n$, in that as $n$ is increased you subtract precisely the reciprocal squared of $n$ each time; I wouldn't have expected increasing the power would have this effect, and there seems to be some obvious connection I've currently not spotted.
 A: Here's a sloppy derivation:
Begin by making the substitution $x=e^u$. We get
\begin{align*}
\int_{1}^{\infty}\frac{\ln(x)}{x^n(x-1)}dx &= \int_{0}^{\infty}\frac{\ln\left(e^u\right)e^u}{\left(e^u\right)^n\left(e^u-1\right)}du\\
&= \int_{0}^{\infty}\frac{u}{e^{nu}e^{-u}\cdot e^{u}\left(1-e^{-u}\right)}du\\
&= \int_{0}^{\infty}\left(\frac{u}{e^{nu}}\sum_{k=0}^{\infty}\left(e^{-u}\right)^k\right)du\\
&= \int_{0}^{\infty}\left(\sum_{k=0}^{\infty}ue^{-ku}e^{-nu}\right)du\\
&= \int_{0}^{\infty}\left(\sum_{k=0}^{\infty}ue^{-(n+k)u}\right)du\\
&= \int_{0}^{\infty}\left(ue^{-nu}+ue^{-(n+1)u}+ue^{-(n+2)u}+\cdots\right)du\\
&= \int_{0}^{\infty}\left(\sum_{k=n}^{\infty}ue^{-ku}\right)du\\
&= \sum_{k=n}^{\infty}\left(\int_{0}^{\infty}ue^{-ku}du\right)\\
&= \sum_{k=n}^{\infty}\left(\frac{1}{k^2}\int_{0}^{\infty}kue^{-ku}\cdot k\text{ }du\right)
\end{align*}
Substituting $v=ku$ and noting that $k\geq n\geq 1$ gives
\begin{align*}
\sum_{k=n}^{\infty}\left(\frac{1}{k^2}\int_{0}^{\infty}kue^{-ku}\cdot k\text{ }du\right) &= \sum_{k=n}^{\infty}\left(\frac{1}{k^2}\int_{0}^{\infty}ve^{-v}dv\right)\\
&= \sum_{k=n}^{\infty}\frac{1}{k^2}
\end{align*}
where the last equality follows from $\int_{0}^{\infty}xe^{-x}dx=1$, a result that can be proven using integration by parts.
The big red flag here is the interchange of the summation and integral. I'm pretty sure there's a theorem that can justify this interchange, but I don't know its name.
