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I have to calculate this integrals

$$I_1 = \int _0^2\:\frac{\arctan \left(x\right)}{x^2+2x+2}dx$$ $$I_2 = \lim _{n\to \infty }\int _0^n\:\frac{\arctan x}{x^2+x+1}dx$$

I have a hint on the first one, the substitution $x = \frac{2-t}{1+2t}$, but I can't understand how it was conceived.

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closed as off-topic by Did, Daniel W. Farlow, user91500, user223391, Namaste Apr 19 '17 at 23:16

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  • $\begingroup$ for $|ab|<1$, $\displaystyle \arctan(a)+\arctan(b)=\arctan\left(\tfrac{a+b}{1-ab}\right)$ $\endgroup$ – FDP Apr 15 '17 at 10:28
  • $\begingroup$ @FDP How can I use this ? $\endgroup$ – Liviu Apr 15 '17 at 10:30
  • $\begingroup$ Perform your change of variable and apply this formula from the right to the left. $\endgroup$ – FDP Apr 15 '17 at 10:34
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    $\begingroup$ if you perform a change of variable to simplify the denominator you would make the argument of arctan(x) more difficult to deal with. But if you could write this arctan stuff as a sum of arctan(x) and arctan(constant) you earn something most of the time. $\endgroup$ – FDP Apr 15 '17 at 10:44
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    $\begingroup$ Other reason is, with your change of variable $\dfrac{1}{x^2+2x+2}dx=-\dfrac{1}{t^2+2t+2}dt$ $\endgroup$ – FDP Apr 15 '17 at 10:53
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$\begin{align} I_2=\int_0^{+\infty} \dfrac{\arctan x}{x^2+x+1}dx=\int_0^{1} \dfrac{\arctan x}{x^2+x+1}dx+\int_1^{+\infty} \dfrac{\arctan x}{x^2+x+1}dx \end{align}$

In the second integral perform the change of variable $y=\dfrac{1}{x}$,

$\begin{align} I_2&=\int_0^{1} \dfrac{\arctan x}{x^2+x+1}dx+\int_0^{1} \dfrac{\arctan\left(\tfrac{1}{x}\right)}{x^2+x+1}dx\\ &=\int_0^{1} \dfrac{\left(\arctan x+\arctan\left(\tfrac{1}{x}\right)\right)}{x^2+x+1}dx\\ &=\int_0^{1} \dfrac{\tfrac{\pi}{2}}{x^2+x+1}dx\\ &=\dfrac{\pi}{2} \int_0^{1} \dfrac{1}{\left(x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}}dx\\ &=\dfrac{\pi}{2}\times \dfrac{2}{\sqrt{3}}\left[\arctan\left(\tfrac{2x+1}{\sqrt{3}}\right)\right]_0^1\\ &=\dfrac{\pi}{\sqrt{3}}\left(\arctan(\sqrt{3})-\arctan\left(\tfrac{1}{\sqrt{3}}\right)\right)\\ &=\dfrac{\pi}{\sqrt{3}}\left(\dfrac{\pi}{3}-\dfrac{\pi}{6})\right)\\ &=\boxed{\dfrac{\pi^2}{6\sqrt{3}}} \end{align}$

Other important relation for arctan function:

If $x>0$ then,

$\arctan x+\arctan\left(\tfrac{1}{x}\right)=\dfrac{\pi}{2}$

If $x<0$ replace $\dfrac{\pi}{2}$ by $-\dfrac{\pi}{2}$

And remember too that, for all $x$ real,

$\arctan(-x)=-\arctan(x)$

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  • $\begingroup$ Thank you very much for the solution and also the the relations for $\arctan$ ! $\endgroup$ – Liviu Apr 15 '17 at 11:41
  • $\begingroup$ i have left behind I_1 that is more harder to deal with but you have a road map to compute it. Just follow it. $\endgroup$ – FDP Apr 15 '17 at 11:47
  • $\begingroup$ Very clever solution. Good job. $\endgroup$ – user12345 Apr 15 '17 at 13:04
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In $I_1$ apply the change of variable $x=\dfrac{2-t}{1+2t}$,

Observe that $t=\dfrac{2-x}{1+2x}$,

$\begin{align}I_1=\int_0^2 \dfrac{\arctan\left(\tfrac{2-t}{1+2t}\right)}{t^2+2t+2}dt \end{align}$

If $t\in [0;2[$ then $0<2t< 1$ therefore,

$\begin{align}I_1=\int_0^2 \dfrac{\arctan(2)}{t^2+2t+2}dt-I_1 \end{align}$

Therefore,

$\begin{align}I_1&=\dfrac{1}{2}\int_0^2 \dfrac{\arctan(2)}{t^2+2t+2}dt\\ &=\dfrac{\arctan(2)}{2}\Big[\arctan\left(x+1\right)\Big]_0^2\\ &=\boxed{\dfrac{\arctan(2)\arctan(3)}{2}-\dfrac{\arctan(2)}{8}\pi} \end{align}$

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  • $\begingroup$ Thank you for solving this too ! This is a multiple choice problem and I have to get to the answer $\frac{1}{2} \arctan 2 \cdot \arctan \frac{1}{2}$, how could I rewrite your answer ? $\endgroup$ – Liviu Apr 15 '17 at 14:19
  • $\begingroup$ Probably you have to omit the last line of my computation of $I_1$ and use the addition formula for arctan i have mentionned above. (i haven't yet tried to get your result) $\endgroup$ – FDP Apr 15 '17 at 14:30
  • $\begingroup$ There is no absolute value in the arctan addition formula the condition is $ab<1$ $\endgroup$ – FDP Apr 15 '17 at 15:08
  • $\begingroup$ Anyway if you had to decide if it's the good unique answer a good approximation is enough to decide. No need to get a closed form/exact value. $\endgroup$ – FDP Apr 15 '17 at 15:30
  • $\begingroup$ Yes, I would have guessed from the choices anyway in the end $\endgroup$ – Liviu Apr 15 '17 at 15:51
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{2}{\arctan\pars{x} \over x^{2} + 2x + 2}\,\dd x & = \Im\int_{0}^{2}{\ln\pars{1 + x\ic} \over x^{2} + 2x + 2}\,\dd x\qquad \pars{~\mbox{set}\ t = 1 + \ic x \implies x = \bracks{1 - t}\ic~} \\[5mm] & = \Re\int_{\ic}^{1 + 2\ic}{\ln\pars{t} \over \pars{t + \ic}\pars{t - 2 + \ic}}\,\dd t \\[5mm] & = {1 \over 2}\,\Re\int_{\ic}^{1 + 2\ic}{\ln\pars{t} \over -\ic - t}\,\dd t - {1 \over 2}\,\Re\int_{\ic}^{1 + 2\ic}{\ln\pars{t} \over 2 - \ic - t}\,\dd t \label{1}\tag{1} \end{align}

However,

\begin{align} \int{\ln\pars{t} \over z - t}\,\dd t & \,\,\,\stackrel{\tau\ =\ t/z}{=}\,\,\, \int{\ln\pars{z\tau} \over 1 - \tau}\,\dd\tau = -\ln\pars{1 - \tau}\ln\pars{z\tau} + \int{\ln\pars{1 - \tau} \over \tau}\,\dd\tau \\[5mm] & = -\ln\pars{1 - \tau}\ln\pars{z\tau} - \,\mrm{Li}_{2}\pars{\tau} = \bbx{\ds{-\ln\pars{1 - {t \over z}}\ln\pars{t} - \,\mrm{Li}_{2}\pars{t \over z}}}\label{2}\tag{2} \end{align}


With \eqref{2}, \eqref{1} becomes \begin{align} &\int_{0}^{2}{\arctan\pars{x} \over x^{2} + 2x + 2}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Re\left\lbrack% -\ln\pars{1 - {1 + 2\ic \over -\ic}}\ln\pars{1 + 2\ic} - \,\mrm{Li}_{2}\pars{1 + 2\ic \over -\ic} + \,\mrm{Li}_{2}\pars{1 \over -\ic}\right. \\[5mm] &\left.\phantom{=} + \ln\pars{1 - {1 + 2\ic \over 2 - \ic}}\ln\pars{1 + 2\ic} + \,\mrm{Li}_{2}\pars{1 + 2\ic \over 2 - \ic} - \mrm{Li}_{2}\pars{1 \over 2 - \ic}\right] \\[1cm] & = {1 \over 8}\bracks{\pi -4\,\mrm{arccot}\pars{3}}\arctan\pars{2} - {1 \over 8}\,\ln^{2}\pars{5} - \,{1 \over 2}\,\Re\mrm{Li}_{2}\pars{-2 + \ic} + {1 \over 2}\,\Re\mrm{Li}_{2}\pars{\ic} \\[5mm] & \phantom{=}+ {1 \over 2}\,\Re\mrm{Li}_{2}\pars{\ic} - {1 \over 2}\,\Re\mrm{Li}_{2}\pars{2 + \ic \over 5} \\[5mm] & = \bbx{\ds{{1 \over 8}\bracks{\pi -4\,\mrm{arccot}\pars{3}}\arctan\pars{2} - {1 \over 8}\,\ln^{2}\pars{5} -\,{\pi^{2} \over 48} - {1 \over 2}\,\Re\mrm{Li}_{2}\pars{-2 + \ic} - {1 \over 2}\,\Re\mrm{Li}_{2}\pars{2 + \ic \over 5}}} \\[5mm] &\ \approx 0.2567 \end{align}

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