I’m trying to compute
$$\lim_{y \to 1-} (1-y + \ln (y))\int_0^y \frac{dx}{(x-1) \ln(x)}$$
I was able to show with Taylor series that this converges to 0, but it was tedious .
Is there a more elegant way to do this perhaps using upper and lower bounds?
Thank you.
Herein, we present a solution that relies only on elementary inequalities and the squeeze theorem. To that end we now proceed.
Let $I(y)$ be the integral given by
$$I(y)=\int_0^y \frac{1}{(x-1)\log(x)}\,dx\tag 1$$
for $y\in (0,1)$.
ASIDE:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1}\tag 2$$
Rearranging the inequalities in $(2)$, we see that for $x<1$
$$\frac{1}{x-1}\le \frac{1}{\log(x)}\le \frac{x}{x-1} \tag3$$
Using $(3)$ in $(1)$, we find that
$$\frac{y}{1-y} +\log(1-y)=\int_0^y \frac{x}{(x-1)^2}\,dx\le I(y)\le \int_0^y \frac{1}{(x-1)^2}\,dx =\frac{y}{1-y} \tag 4$$
Multiplying $(4)$ by $(1-y+\log(y))$ reveals
$$(1-y+\log(y))\left(\frac{y}{1-y} +\log(1-y)\right)\le (1-y+\log(y))I(y)\le (1-y+\log(y))\left(\frac{y}{1-y}\right)\tag 5$$
whereupon applying the squeeze theorem to $(5)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 1^-}(1-y+\log(y))\int_0^y \frac{1}{(x-1)\log(x)}\,dx=0}$$
Substitute $y=1-t$, so the limit becomes $$ \lim_{t\to0^+}(t+\ln(1-t))\int_0^{1-t}\frac{1}{(x-1)\ln x}\,dx= \lim_{t\to0^+}-\frac{t^2}{2}\int_0^{1-t}\frac{1}{(x-1)\ln x}\,dx $$ because $$ \lim_{t\to0^+}\frac{t+\ln(1-t)}{t^2}=-\frac{1}{2} $$ In the integral, do the substitution $x=1-u$, so we get $$ \lim_{t\to0^+}\frac{t^2}{2}\int_t^1\frac{1}{u\ln(1-u)}\,du $$ Now change $t=1/s$, so the limit is $$ \lim_{s\to\infty} \frac{\displaystyle\int_{1/s}^1\dfrac{1}{u\ln(1-u)}\,du}{2s^2}= \lim_{s\to\infty}\dfrac{\dfrac{1}{s^2}\dfrac{1}{(1/s)\ln(1-1/s)}}{4s}= \lim_{t\to0^+}\frac{t^2}{4\ln(1-t)}=0 $$
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\lim_{y \to 1^{\large-}}\braces{\bracks{1 - y + \ln\pars{y}} \int_{0}^{y}{\dd x \over \pars{x - 1}\ln\pars{x}}}:\ {\large ?}}$.
\begin{align} &\int_{0}^{y}{\dd x \over \pars{x - 1}\ln\pars{x}} = \int_{0}^{y}\bracks{{1 \over \pars{x - 1}\ln\pars{x}} - {1 \over \pars{x - 1}^{2}} - {1 \over 2\pars{x - 1}}}\dd x \\[2mm] + &\ \int_{0}^{y}\bracks{ {1 \over \pars{x - 1}^{2}} + {1 \over 2\pars{x - 1}}}\dd x \\[1cm] = &\ \int_{0}^{y}\bracks{{1 \over \pars{x - 1}\ln\pars{x}} - {1 \over \pars{x - 1}^{2}} - {1 \over 2\pars{x - 1}}}\dd x \\[2mm] - &\ {1 \over y - 1} - 1 + {1 \over 2}\ln\pars{1 - y} \sim c - {1 \over y - 1} + {1 \over 2}\ln\pars{1 - y}\quad \mbox{as}\ y \to 1^{\large -}\,,\ \pars{~c:\ constant~} \end{align}
and $\ds{1 - y + \ln\pars{y} \sim -\,{1 \over 2}\pars{y - 1}^{2} \quad\mbox{as}\ y \to 1^{\large -}}$.
