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I tried with partial decomposition and no...

$$ \int_0^\infty\frac{x^2}{1+x^{10}}\;dx = \frac{\pi}{5\phi}. $$

Gary
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TeSan
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  • It looks like $1+x^2$ is a factor of the denominator... Can you list the steps you took with partial fractions? – abiessu Sep 28 '22 at 17:57
  • Partial fraction is helping...im posting the indefinite integration...put the limit yourself – Vanessa Sep 28 '22 at 17:59
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    $$ \int_0^{ + \infty } {\frac{{x^2 }}{{1 + x^{10} }}dx} = \frac{1}{{10}}\int_0^{ + \infty } {\frac{{t^{3/10 - 1} }}{{1 + t}}dt} = \frac{1}{{10}}B\left( {\tfrac{3}{{10}},1 - \tfrac{3}{{10}}} \right) \ = \frac{1}{{10}}\Gamma \left( {\tfrac{3}{{10}}} \right)\Gamma \left( {1 - \tfrac{3}{{10}}} \right) = \frac{\pi }{{10\sin \left( {\frac{{3\pi }}{{10}}} \right)}}=\frac{\pi}{5\phi} $$ – Gary Sep 28 '22 at 18:14
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    This can be calculated in a pretty standard way using the residue theorem and a half-circle contour integral in the complex plane. I can show the details if you are interested – Andrew Sep 28 '22 at 19:31

2 Answers2

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This is just an expression not an answer

$$\dfrac{\sqrt{2}\ln\left(\sqrt{2}x^2+\sqrt{\sqrt{5}+5}x+\sqrt{2}\right)-\sqrt{2}\ln\left(\sqrt{2}x^2-\sqrt{\sqrt{5}+5}x+\sqrt{2}\right)}{2\sqrt{16{\cdot}5^\frac{3}{2}+200}-2\left(5-\sqrt{5}\right)\sqrt{\sqrt{5}+5}}+\dfrac{\sqrt{2}\ln\left(\sqrt{2}x^2-\sqrt{5-\sqrt{5}}x+\sqrt{2}\right)-\sqrt{2}\ln\left(\sqrt{2}x^2+\sqrt{5-\sqrt{5}}x+\sqrt{2}\right)}{2\sqrt{5-\sqrt{5}}\left(\sqrt{5}+5\right)-2\sqrt{200-16{\cdot}5^\frac{3}{2}}}+\dfrac{2^\frac{7}{4}\left(\sqrt{2{\cdot}5^\frac{3}{2}+25}-\frac{5\sqrt{\sqrt{5}+5}}{2^\frac{3}{2}}\right)\arctan\left(\frac{2^\frac{3}{2}x+\sqrt{\sqrt{5}+5}}{2^\frac{5}{4}\sqrt{\sqrt{2}-\frac{\sqrt{5}+5}{2^\frac{5}{2}}}}\right)-2^\frac{7}{4}\left(\frac{5\sqrt{\sqrt{5}+5}}{2^\frac{3}{2}}-\sqrt{2{\cdot}5^\frac{3}{2}+25}\right)\arctan\left(\frac{2^\frac{3}{2}x-\sqrt{\sqrt{5}+5}}{2^\frac{5}{4}\sqrt{\sqrt{2}-\frac{\sqrt{5}+5}{2^\frac{5}{2}}}}\right)}{5\sqrt{\sqrt{2}-\frac{\sqrt{5}+5}{2^\frac{5}{2}}}\left(2\sqrt{16{\cdot}5^\frac{3}{2}+200}-2\left(5-\sqrt{5}\right)\sqrt{\sqrt{5}+5}\right)}+\dfrac{2^\frac{7}{4}\left(\frac{5\sqrt{5-\sqrt{5}}}{2^\frac{3}{2}}-\sqrt{25-2{\cdot}5^\frac{3}{2}}\right)\arctan\left(\frac{2^\frac{3}{2}x-\sqrt{5-\sqrt{5}}}{2^\frac{5}{4}\sqrt{\frac{\sqrt{5}-5}{2^\frac{5}{2}}+\sqrt{2}}}\right)-2^\frac{7}{4}\left(\sqrt{25-2{\cdot}5^\frac{3}{2}}-\frac{5\sqrt{5-\sqrt{5}}}{2^\frac{3}{2}}\right)\arctan\left(\frac{2^\frac{3}{2}x+\sqrt{5-\sqrt{5}}}{2^\frac{5}{4}\sqrt{\frac{\sqrt{5}-5}{2^\frac{5}{2}}+\sqrt{2}}}\right)}{5\sqrt{\frac{\sqrt{5}-5}{2^\frac{5}{2}}+\sqrt{2}}\left(2\sqrt{5-\sqrt{5}}\left(\sqrt{5}+5\right)-2\sqrt{200-16{\cdot}5^\frac{3}{2}}\right)}-\dfrac{\arctan\left(x\right)}{5}$$ After puttting limits we obtain the expression $$\dfrac{{\pi}}{10\sin\left(\frac{3{\pi}}{10}\right)}$$

Vanessa
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1

The antiderivative is not difficult using partial fraction decomposition.

Write $$\frac {x^2}{x^{10}+1}=\frac {x^2}{(x^{2}+1)(x^2-a)(x^2-b)(x^2-c)(x^2-d)}$$ where $(a,b,c,d)$ are the known complex solutions of the quartic $$x^4-x^3+x^2-x+1=0$$

Claude Leibovici
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