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How to prove the following result?

$$\int_0^1\frac{x^3\arctan x}{(3-x^2)^2}\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{\pi\sqrt{2}}{192}\left(18-\pi-6\sqrt{3}\,\right)$$

For my part no idea?

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  • $\begingroup$ My long shot is to use $\arctan x=\Im\log(1+ix)$ then use differentiation under the integral sign. But it might lead to nowhere. $\endgroup$ Jan 12, 2015 at 23:50
  • $\begingroup$ I'm leaving a comment here so I keep receiving notifications on this one. I am very curious what the solution pans out to $\endgroup$ Jan 12, 2015 at 23:57
  • $\begingroup$ An interesting fact (despite the fact that I cheated by using WA) is that $$\int\frac{x^3}{(3-x^2)^2\sqrt{1-x^2}}\,dx = \frac{1}{4\sqrt{2}}\arctan\sqrt{\frac{1-x^2}{2}}-\frac{3}{4}\frac{\sqrt{1-x^2}}{3-x^2},$$ so integration by parts followed by the residue theorem may be the way to go. $\endgroup$ Jan 13, 2015 at 0:30
  • $\begingroup$ I will post my answer as soon as possible. The idea is to introduce two parameters. $\endgroup$
    – user111187
    Jan 13, 2015 at 9:13
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    $\begingroup$ @frogeyedpeas it has been solved $\endgroup$
    – RE60K
    Jan 13, 2015 at 14:07

4 Answers 4

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Although integration by parts will work, a nice general way to attack the integral is by introducing

$$ J(a,b) := \int_0^1 \frac{x \arctan(ax)}{1-b^2 x^2} \frac{dx}{\sqrt{1-x^2}}, $$ so $J(0,b) = 0$ and $$ \begin{align} \partial_a J &= \int_0^1 \frac{x^2}{(1-b^2 x^2)(1+a^2x^2)} \frac{dx}{\sqrt{1-x^2}} = \int_0^{\pi/2} \frac{du\sec^2 u}{(\sec^2 u -b^2)(\sec^2 u +a^2)} \\&= \int_0^{\infty} \frac{dt}{(t^2 + 1-b^2)(t^2 + 1+a^2)} = \frac{\pi/2}{\sqrt{1-b^2}\sqrt{1+a^2}} \frac{1}{\sqrt{1-b^2}+\sqrt{1+a^2}}, \end{align} $$ where $ x = \cos u, t = \tan u$, and the last equality follows easily using partial fractions.

Hence $$ \begin{align} J(1,b) &= \frac{\pi/2}{\sqrt{1-b^2}} \int_0^1 \frac{da}{\sqrt{1+a^2}} \frac{1}{\sqrt{1-b^2}+\sqrt{1+a^2}} \\&= \left.\frac{\pi/2}{\sqrt{1-b^2}} \frac{1}{b} \left[\arctan\left(\frac{b}{\sqrt{1-b^2}}\frac{\sqrt{1+a^2}}{a} \right)- \arctan \frac b a\right] \right\lvert_{a=0}^{a=1} \\&= \frac{\pi/2}{\sqrt{1-b^2}} \frac{1}{b} \left[\arctan\sqrt\frac{2b^2}{1-b^2}- \arctan b\right]. \end{align} $$

This implies that $$ \begin{align} &2b^{-3} \int_0^1 \frac{x^3 \arctan(x)}{(b^{-2} - x^2)^2} \frac{dx}{\sqrt{1-x^2}} = \partial_b J(1,b) \\&= \frac{\pi}{2b^2(1-b^2)^{3/2}} \left\{b\frac{ \sqrt 2\sqrt{1- b^2}+2}{1+b^2}+\left(1-2b^2\right) \left[\arctan b-\arctan \sqrt{\frac{2b^2}{1-b^2}}\right]-b\right\}. \end{align} $$

Plugging in $b = \dfrac{1}{\sqrt 3}$ simplifies this a lot, giving $$ \int_0^1 \frac{x^3 \arctan(x)}{(3 - x^2)^2} \frac{dx}{\sqrt{1-x^2}} =\frac{\pi \sqrt 2}{192}\left[18-\pi - 6\sqrt 3 \right], $$ as desired.

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  • $\begingroup$ This is a good answer & I upvote it but it would be nice if you mention what kind of substitutions did you use to evaluate the integral. $$x\stackrel{?}\to u\stackrel{?}\to t$$ $\endgroup$
    – Venus
    Jan 13, 2015 at 10:13
  • $\begingroup$ Magnificient proof, tank you infinitely(+1) $\endgroup$
    – user178256
    Jan 13, 2015 at 19:23
  • $\begingroup$ @user178256 Please consider accepting the answer if it is to your satisfaction. $\endgroup$
    – user111187
    Jan 13, 2015 at 19:43
  • $\begingroup$ very nice solution,very satisfied with your answer $\endgroup$
    – user178256
    Jan 13, 2015 at 23:18
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$$\newcommand{\b}[1]{\left(#1\right)} \newcommand{\d}{{\rm d}} \newcommand{\f}{\frac} \newcommand{\s}{\sqrt} \newcommand{\t}{\text} \newcommand{\u}{\underbrace} \bf Answer$$

This belongs to a large category of functions that are easily solvable with integration by parts like:

$$\int\frac{x\arctan(x)}{\sqrt{1+x^2}}\d x=\sqrt{1+x^2}\arctan(x)-\ln|x+\sqrt{x^2+1}|+c$$


So let our functions be: $$f(x)=\f{x^3}{(3-x^2)^2\s{1-x^2}};g(x)=\arctan(x)\\\int f(x)g(x)\d x=g(x)\int f(x)\d x-\int g'(x)\u{\int f(x)\d x}_{h(x)}\d x$$ And(Beware I used Reduction Formula for evaluating last integral!): $$\int f(x)\d x=\f12\int\f{t\d t}{(3-t)^2\s{1-t}}\quad(t=x^2)\\ =\int\f{u^2-1}{(u^2+2)^2}\d u=\int\f1{(u^2+2)}\d u-3\int\f{1}{(u^2+2)^2}\d u\quad(u^2=1-t)\\ =\frac1{\sqrt2}\arctan\frac u{\sqrt2}-3\b{\frac u{4(u^2+2)}+\f1{4\s2}\arctan\f u{\s2}}+{\cal C}\\ =\f1{4\s2}\arctan\s{\f{1-x^2}2}-\f34\f{\s{1-x^2}}{3-x^2}+{\cal C}=\t{(let) }h(x)$$ And thus: $$\int g'(x)h(x)\d x=\f1{4\s2}\int \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x-\f34\int \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x$$ So:

$$I=\int_0^1f(x)g(x)\d x=\b{\arctan (x)\b{\f1{4\s2}\arctan\s{\f{1-x^2}2}-\f34\f{\s{1-x^2}}{3-x^2}}}_0^1\\-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x+\f34\int_0^1 \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x\\ I=\f34\u{\int_0^1 \f{\s{1-x^2}}{(3-x^2)(1+x^2)}\d x}_{I_1}-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x$$

Now: $$I_1=\int_0^\infty\f{\d y}{4y^4+8y^2+3}\quad\b{x=\f{y^2}{y^2+1}\t{ or }y=\f{x}{\s{1-x^2}}=\tan\arcsin x \\\t{ which can also be obtained after two sunstitutions viz., }x=\sin z\t{, then }y=\tan z}\\ \f12\b{\f1{2y^2+1}-\f1{2y^2+3}}_0^{\infty}\\ =\f{3-\s3}{24}\pi\s2=\f{3-\s3}{24}\pi\s2$$

Still working on the last integral!! BTW the final form is: $$I=\frac{\pi\sqrt{2}}{192}(18-6\sqrt{3})-\f1{4\s2}\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x$$ So it remains to prove: $$\int_0^1 \f1{1+x^2}\arctan\s{\f{1-x^2}2}\d x=\pi^2/24$$

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    $\begingroup$ This is actually the approach I took at first. To evaluate the last integral, I integrated by parts and got something like $\int dx \frac{x \arctan x}{(3-x^2)\sqrt{1-x^2}}$. It is not really easier than calculating the general case $J(a,b)$. $\endgroup$
    – user111187
    Jan 13, 2015 at 17:19
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}\frac{x^{3}\arctan\pars{x}}{\pars{3 - x^{2}}^{2}} \frac{\dd x}{\root{1 - x^{2}}}=\frac{\pi\root{2}}{192}\, \pars{18 - \pi - 6\root{3}} \approx {\tt 0.1033}:\ {\large ?}}$


\begin{align}&\overbrace{\color{#66f}{\large% \int_{0}^{1}\frac{x^{3}\arctan\pars{x}}{\pars{3 - x^{2}}^2} \frac{\dd x}{\root{1 - x^{2}}}}}^{\ds{\dsc{x}\ \mapsto\ \dsc{\frac{1}{x}}}}\ =\ \int_{1}^{\infty}\frac{\arctan\pars{1/x}}{\pars{3x^{2} - 1}^{2}} \frac{\dd x}{\root{x^{2} - 1}} \\[5mm]&=\int_{1}^{\infty}\ \overbrace{\int_{0}^{1}\frac{x\,\dd t}{t^{2} + x^{2}}} ^{\dsc{\arctan\pars{1/x}}}\ \frac{\dd x}{\pars{3x^{2} - 1}^{2}\root{x^{2} - 1}} =\int_{0}^{1}\ \overbrace{\int_{1}^{\infty} \frac{x\,\dd x}{\pars{x^{2} + t^{2}}\pars{3x^{2} - 1}^{2}\root{x^{2} - 1}}} ^{\ds{\dsc{x}\ \mapsto\ \dsc{\root{x}}}}\ \,\dd t \\[5mm]&=\half\int_{0}^{1}\int_{1}^{\infty} \frac{\dd x}{\pars{x + t^{2}}\pars{3x - 1}^{2}\root{x - 1}}\,\dd t \\[5mm]&=\left.\frac{1}{18}\,\totald{}{\mu}\int_{0}^{1}\int_{1}^{\infty} \frac{\dd x}{\pars{x + t^{2}}\pars{x - \mu}\root{x - 1}}\,\dd t \right\vert_{\mu\ =\ 1/3} \\[5mm]&=\left.\frac{1}{18}\,\totald{}{\mu}\int_{0}^{1} \frac{1}{t^{2} + \mu}\braces{% \underbrace{\int_{1}^{\infty}\frac{\dd x}{\pars{x - \mu}\root{x - 1}}} _{\dsc{\frac{\pi}{\root{1 - \mu}}}}\ -\ \underbrace{\int_{1}^{\infty}\frac{\dd x}{\pars{x + t^{2}}\root{x - 1}}} _{\dsc{\frac{\pi}{\root{t^{2} + 1}}}}}\,\dd t \right\vert_{\mu\ =\ 1/3} \end{align} The '$\ds{x}$-integrations' were easily evaluated with the change $\ds{\dsc{\root{x - 1}}\ \mapsto\ \dsc{x}}$. Note that $\ds{\int_{0}^{1}\frac{\dd t}{t^{2} + \mu} =\frac{\arctan\pars{\mu^{-1/2}}}{\root{\mu}}}$. Namely, \begin{align}&\color{#66f}{\large% \int_{0}^{1}\frac{x^{3}\arctan\pars{x}}{\pars{3 - x^{2}}^2} \frac{\dd x}{\root{1 - x^{2}}}} \\[5mm]&=\overbrace{\frac{\pi}{18}\,\totald{}{\mu}\bracks{% \frac{\arctan\pars{\mu^{-1/2}}}{\root{\mu\pars{1 - \mu}}}}_{\mu\ =\ 1/3}} ^{\dsc{-\,\frac{\pi\root{2}}{192}\,\pars{4\pi + 6\root{3}}}}\ -\,\,\, \overbrace{\left.\frac{\pi}{18}\,\totald{}{\mu} \int_{0}^{1}\frac{\dd t}{\pars{t^{2} + \mu}\root{t^{2} + 1}} \right\vert_{\mu\ =\ 1/3}} ^{\dsc{-\,\frac{\pi\root{2}}{192}\,\pars{3\pi + 18}}} \\[5mm]&=\color{#66f}{\large\frac{\pi\root{2}}{192}\, \pars{18 - \pi - 6\root{3}}} \approx {\tt 0.1033} \end{align} The last integral is evaluated as follows: \begin{align}&\overbrace{% \int_{0}^{1}\frac{\dd t}{\pars{t^{2} + \mu}\root{t^{2} + 1}}} ^{\ds{\dsc{t}\ \mapsto\ \dsc{\frac{1}{t}}}} =\int_{1}^{\infty}\frac{t\,\dd t}{\pars{1 + \mu t^{2}}\root{t^{2} + 1}} =\half\int_{1}^{\infty}\frac{\dd t}{\pars{1 + \mu t}\root{t + 1}} \\[5mm]&=\frac{1}{2\mu} \int_{2}^{\infty}\frac{\dd t}{\pars{t - 1 + 1/\mu}\root{t}} =\frac{1}{\mu} \int_{\root{2}}^{\infty}\frac{\dd t}{t^{2} + 1/\mu - 1} =\frac{\arctan\pars{\sqrt{\frac{1}{2\mu}-\half}}} {\sqrt{\mu\pars{1 - \mu}}} \end{align}

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  • $\begingroup$ Tank for the answer. $\endgroup$
    – user178256
    Jan 31, 2015 at 22:35
  • $\begingroup$ @user178256 I'm glad it was useful for you. Thanks. $\endgroup$ Feb 9, 2015 at 19:00
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$\bf{Modified:}$

Based on an idea from the solution of @user111187:

Consider a general integral of the form

$$I_b =\int_0^1 \frac{R(x)}{\sqrt{1-x^2}} \arctan(b x)dx$$

with derivative $$J_b \colon =\frac{dI_b}{db} = \int_0^1 \frac{x\, R(x)}{(1+b^2 x^2)\sqrt{1-x^2}} dx$$ and $I_0=0$.

The integral $J_b$ is doable if $R$ is a rational function. We get

$$I =I_1 = \int_{0}^1 J_b\, db$$

This last integral is doable in some cases.

In our case $R(x) = \frac{x^3}{(3-x^2)^2}$ so we get

$$J_b = \int_0^1 \frac{x^4}{(1+b^2 x^2)(3-x^2)^2 \sqrt{1-x^2}} dx$$

We find first the indefinite integral

$$\int \frac{x^4}{(1+b^2 x^2)(3-x^2)^2 \sqrt{1-x^2}} dx $$ One can check that it equals $$ F_b(x):=\frac{3 x \sqrt{1-x^2}}{4 \left(1+3 b^2\right) \left(-3+x^2\right)}+\frac{3 \sqrt{\frac{3}{2}} \left(-1+b^2\right) \text{ArcTan}\left[\frac{\sqrt{\frac{2}{3}} x}{\sqrt{1-x^2}}\right]}{4 \left(1+3 b^2\right)^2}+\frac{\text{ArcTan}\left[\frac{\sqrt{1+b^2} x}{\sqrt{1-x^2}}\right]}{\sqrt{1+b^2} \left(1+3 b^2\right)^2}$$ Indeed, by direct calculations $$F_b'(x) = \frac{x^4}{(1+b^2 x^2)(3-x^2)^2 \sqrt{1-x^2}}$$

We get $$J_b = F_b(1_{-})- F_b(0) = \frac{\left(8-3 \sqrt{6}(1+b^2)^{\frac{3}{2}}\right) \pi }{16 \sqrt{1+b^2} \left(1+3 b^2\right)^2}$$

and therefore $$I = I_1 = \int_0^1 \frac{\left(8-3 \sqrt{6}(1+b^2)^{\frac{3}{2}}\right) \pi }{16 \sqrt{1+b^2} \left(1+3 b^2\right)^2} db = \frac{\pi \sqrt{2}}{192}\cdot (\,18 - 6 \sqrt{3} - \pi) $$

This approach works for all integrals of form

$$\int_0^1 \frac{R(x)}{\sqrt{1-x^2}} \arctan(b x)dx=\int_0^1 \frac{x \, Q(x^2)}{\sqrt{1-x^2}} \arctan(x) dx$$

where $R(x) = xQ(x^2)$ is an odd rational function.

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  • $\begingroup$ Tank for the answer $\endgroup$
    – user178256
    Jan 31, 2015 at 22:36

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