Integral limit without using Taylor expansion 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.

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
$$
\lim_{s\to\infty}
\frac{\displaystyle\int_{1/s}^1\dfrac{1}{u\ln(1-u)}\,du}{2s^2}=
\lim_{s\to\infty}\dfrac{\dfrac{1}{s^2}\dfrac{1}{(1/s)\ln(1-1/s)}}{4s}=
\lim_{t\to0^+}\frac{t^2}{4\ln(1-t)}=0
$$
