Limit of $\lim_{t\to 1^+} \int_t^{t^2} \frac{\arctan(x)}{x-1} dx$

Question

I need to evaluate $$\lim_{t\to 1^+} \int_t^{t^2} \frac{\arctan(x)}{x-1} \; dx.$$ I have tried the partial integrating, and I got confused. Could someone help me, please?

Answer 1

$$\arctan(t^2)\ln(\frac{t^2-1}{t-1})\geq\int\limits_t^{t^2} \frac{\arctan(x)}{x-1}dx\geq\arctan(t)\ln(\frac{t^2-1}{t-1})$$ By squeeze theorem, the limit is $\frac{\pi}{4}\ln(2)$

Answer 2

$artan (x) \to \frac {\pi} 4$ as $ x \to 1$. Hence the lmit is $ \frac {\pi} 4 \lim \int_t^{t^{2}} \frac 1 {x-1} \, dt=(log \, 2 ) \frac {\pi} 4 $

Answer 3

Write $t = 1+\epsilon$ and notice that

\begin{align*} \int_{t}^{t^2} \frac{\arctan x}{x-1} \, dx &\stackrel{(x=1+\epsilon u)}{=} \int_{1}^{2+\epsilon} \frac{\arctan(1+\epsilon u)}{u} \, du \\ &\xrightarrow[\epsilon\downarrow0]{} \int_{1}^{2} \frac{\arctan(1)}{u} \, du = \frac{\pi}{4}\log 2. \end{align*}

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{t \to 1^{\large +}}\int_{t}^{t^{2}} {\arctan\pars{x} \over x - 1}\,\dd x} \\[5mm] = &\ \lim_{t \to 1^{\large +}}\int_{t}^{t^{2}} {\arctan\pars{x} - \arctan\pars{1} \over x - 1}\,\dd x\ +\ \lim_{t \to 1^{\large +}}\underbrace{\int_{t}^{t^{2}} {\arctan\pars{1} \over x - 1}\,\dd x} _{\ds{=\ {\pi \over 4}\,\ln\pars{t + 1}}}\label{1}\tag{1} \end{align}

Note that $\ds{\left.{\arctan\pars{x} - \arctan\pars{1} \over x - 1} \,\right\vert_{\ 1\ <\ t\ <\ x\ <\ t^{2}} = \left.{1 \over \xi^{2} + 1} \,\right\vert_{\ 1 <\ \xi\ <\ x}}$

$\ds{\implies \bbx{{1 \over x^{2} + 1} < \left.{\arctan\pars{x} - \arctan\pars{1} \over x - 1} \,\right\vert_{\ 1\ <\ t\ <\ x\ <\ t^{2}} < {1 \over 2}}}$

The first limit in \eqref{1} vanishes out because $$ \arctan\pars{t^{2}} - \arctan\pars{t} < \int_{t}^{t^{2}} {\arctan\pars{x} - \arctan\pars{1} \over x - 1}\,\dd x < {t\pars{t - 1} \over 2} $$ such that \eqref{1} becomes $$ \bbox[10px,#ffd]{\lim_{t \to 1^{\large +}}\int_{t}^{t^{2}} {\arctan\pars{x} \over x - 1}\,\dd x} = \bbx{{\pi \over 4}\,\ln\pars{2}} $$