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Evaluation of $$\int_\limits{-1}^{3}\frac{\arctan(1+x^2)}{x}dx$$

$\bf{My\; Try::}$ Let $$I = \int_\limits{-1}^3\arctan(1+x^2)\cdot \frac{1}{x}dx$$

Using By parts, We get

$$I = \left[\arctan (1+x^2)\cdot \ln|x|\right]^{3}_{-1}-2\int_\limits{-1}^3\frac{x\ln |x|}{1+x^4}dx$$

Now How can i solve it , Help required, Thanks

Jose M Serra
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juantheron
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    Maybe, you are looking for the $Principal\ Value$. Otherwise, it diverges ' logarithmically around' $x = 0$ as $\texttt{@Jacky Chong}$ already pointed out in his answer. – Felix Marin Oct 31 '16 at 06:16
  • that's how the principalvalue integral looks: $$ \frac{1}{4} \left(\pi \log (3)+i \left(\text{Li}_2\left(-\frac{1}{2}+\frac{i}{2}\right)-\text{Li}_2\left(-\frac{9}{2}+\frac{9 i}{2}\right)-\text{Li}_2\left(-\frac{1}{2}-\frac{i}{2}\right)+\text{Li}_2\left(-\frac{9}{2}-\frac{9 i}{2}\right)\right)\right) $$ – tired Oct 31 '16 at 15:19

4 Answers4

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By continuity, there exists a neighborhood about zero such that \begin{align} |\arctan 1 -\arctan (1+x^2)| < \frac{1}{4} \ \ \Rightarrow \ \ \arctan1<\frac{1}{4}+\arctan (1+x^2) \end{align} which means \begin{align} 0<\frac{\frac{\pi}{4}-\frac{1}{4}}{x} < \frac{\arctan(1+x^2)}{x} \end{align} whenever $r<x<0$ for some $r$. Now, observe \begin{align} \infty =\int^0_{-1} \frac{\frac{\pi-1}{4}}{x}\ dx< \int^0_{-1} \frac{\arctan(1+x^2)}{x}\ dx. \end{align}

Jacky Chong
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$$\arctan(x^2+1) > \arctan (1)=\frac{\pi}{4}$$

For $x \in (0,\infty)$ as $x^2 > 0 \implies x^2+1 > 1$ and $f(u)=\arctan(u)$ is strictly increasing on $x \in (0,\infty)$.

Hence,

$$\int_{0}^{3} \frac{\arctan (x^2+1)}{x} > \int_{0}^{3} \frac{\pi}{4} \frac{1}{x} \to \infty$$

Ahmed S. Attaalla
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The integral does not converge as has been explained above, however it does have a finite principal value (apparently PV is $\frac{31423231}{21789317}$ or 1.44214 )

user140776
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The original integral clearly diverges logarithmically "at $\ds{x = 0}$". We'll evaluate the Principal Value by assuming it was the OP original intention.

\begin{align} &\mrm{P.V.}\int_{-1}^{3}{\arctan\pars{1 + x^{2}} \over x}\,\dd x \\[5mm] = &\ \lim_{\epsilon \to 0^{+}}\bracks{% \int_{-1}^{-\epsilon}{\arctan\pars{1 + x^{2}} \over x}\,\dd x + \int_{\epsilon}^{3}{\arctan\pars{1 + x^{2}} \over x}\,\dd x} \\[5mm] = &\ \ \overbrace{\mrm{P.V.}\int_{-1}^{1}{\arctan\pars{1 + x^{2}} \over x}\,\dd x} ^{\ds{=\ 0}}\ +\ \int_{1}^{3}{\arctan\pars{1 + x^{2}} \over x}\,\dd x\quad \pars{\begin{array}{c} \mbox{Integrate by parts}\\ \mbox{the last integral} \end{array}} \\[5mm] = &\ \ln\pars{3}\arctan\pars{10} - \int_{1}^{3}\ln\pars{x}\,{2x \over \pars{1 + x^{2}}^{2} + 1}\,\dd x \\[5mm] \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=} &\,\,\, \ln\pars{3}\arctan\pars{10} - {1 \over 2}\int_{1}^{9}{\ln\pars{x} \over \pars{1 + x}^{2} + 1}\,\dd x \\[5mm] = &\ \ln\pars{3}\arctan\pars{10} - {1 \over 2}\int_{1}^{9}\ln\pars{x} \pars{{1 \over x + 1 - \ic} - {1 \over x + 1 + \ic}}{1 \over 2\ic}\,\dd x \\[5mm] = &\ \ln\pars{3}\arctan\pars{10} + {1 \over 2}\,\Im\int_{1}^{9}{\ln\pars{x} \over -1 + \ic - x}\,\dd x \\[5mm] = &\ \ln\pars{3}\arctan\pars{10} + {1 \over 2}\,\Im\int_{1}^{9}{% \ln\pars{\bracks{-1 + \ic}\braces{x/\bracks{-1 + \ic}}} \over 1 - x/\pars{-1 + \ic}}\,{\dd x \over -1 + \ic} \\[5mm] \stackrel{x/\pars{-1 + \ic}\ \mapsto\ x}{=} &\,\,\, \ln\pars{3}\arctan\pars{10} + {1 \over 2}\,\Im\int_{-\pars{1 + \ic}/2}^{-9\pars{1 + \ic}/2}{% \ln\pars{\bracks{-1 + \ic}x} \over 1 - x}\,\dd x \\[1cm] = &\ \ln\pars{3}\arctan\pars{10} + {1 \over 2}\,\Im\left\lbrace% -\ln\pars{1 + {9 \over 2}\,\bracks{1 + \ic}}\ln\pars{9}\right. \\[5mm] &\phantom{\ln\pars{3}\arctan\pars{10} + {1 \over 2}\,\Im\braces{}} + \left.\int_{-\pars{1 + \ic}/2}^{-9\pars{1 + \ic}/2}{% \ln\pars{1 - x} \over x}\,\dd x\right\rbrace \\[1cm] = &\ \ln\pars{3}\arctan\pars{10} - \ln\pars{3} \arctan\pars{9 \over 11} - {1 \over 2}\,\Im \int_{-\pars{1 + \ic}/2}^{-9\pars{1 + \ic}/2} \mrm{Li}_{2}'\pars{x}\,\dd x \\[5mm] = &\ \bbox[#ffe,15px,border:1px dotted navy]{\ds{{1 \over 4}\,\ln\pars{3}\,\pi - {1 \over 2}\,\Im\braces{% \mrm{Li}_{2}\pars{-\,{9 \over 2}\,\bracks{1 + \ic}} - \mrm{Li}_{2}\pars{-\,{1 + \ic \over 2}}}}}\ \approx 1.4421 \end{align}

Note that

$\ds{\arctan\pars{10} - \arctan\pars{9/11} = \arctan\pars{\bracks{10 - 9/11}/\braces{1 + 10\bracks{9/11}}} = \arctan\pars{1} = \pi/4}$.

Felix Marin
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