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$$\displaystyle\int_{0}^{\infty}{1\over x^8+x^4+1}\cdot{1\over x^3+1}\mathrm dx={5\pi\over 12\sqrt{3}}$$

$x^8+x^4+1=(x^2+x-1)(x^2-x-1)(x^2+x+1)(x^2-x+1)$

$x^3+1=(x+1)(x^2-x+1)$

${A\over x+1}+{Bx+C\over x^2-x+1}={1\over x^3+1}$

${Ax+B\over x^2+x-1}+{Cx+D\over x^2-x-1}+{Ex+F\over x^2+x+1}+{Gx+H\over x^2-x+1}={1\over x^8+x^4+1}$

it would be a nightmare trying to decomposition of fraction.

I would like some help please?

  • 2
    Do you know complex integration? This might be solved nicely by the Residue Theorem. – 2017-01-08
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    You may avoid the partial fraction decomposition by carefully exploiting https://en.wikipedia.org/wiki/Digamma_function#Reflection_formula – 2017-01-08
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    I solved with Maple but the result was ${\frac {5}{36}}\,\sqrt {3}\pi$. – 2017-01-08
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    @Amin235: that is the same as $\frac{5\pi}{12\sqrt{3}}$. – 2017-01-08
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    You right, I really was confused that made this comment. – 2017-01-09
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    @TheCount It doesn't really lend itself to solving with the residue theorem. The integral is from 0 to infinity, the integrand is not even, and there are no branch cuts. And I don't see any weird contour that just makes it work, either. – 2017-01-09
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    @Chris, well it was just a quick thought. No guarantees! Ha. – 2017-01-09

2 Answers 2

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We want to compute: $$ I = \int_{0}^{1}\frac{x^4-1}{x^{12}-1}\cdot\frac{\mathrm dx}{x^3+1}+\int_{0}^{1}\frac{x^{8}-x^{12}}{1-x^{12}}\cdot\frac{x\,\mathrm dx}{x^3+1}=\int_{0}^{1}\frac{1-x^3+x^6}{1+x^4+x^8}\,\mathrm dx $$ that is: $$ I = \int_{0}^{1}\left(1-x^3-x^4+x^6+x^7-x^{10}\right)\frac{\mathrm dx}{1-x^{12}} $$ or: $$ I=\sum_{k\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+4}-\frac{1}{12k+5}+\frac{1}{12k+7}+\frac{1}{12k+8}-\frac{1}{12k+11}\right). $$ By the reflection formula for the digamma function we have: $$ \sum_{k\geq 0}\left(\frac{1}{12k+\tau}-\frac{1}{12k+(12-\tau)}\right)=\frac{\pi}{12}\,\cot\frac{\pi \tau}{12} $$ and the wanted result trivially follows from $\cot\left(\frac{\pi}{12}\right)=2+\sqrt{3}$.

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    @DejanGovc: there might be a sign mistake but my final result match yours, so it shouldn't be. I splitted the integration range as $(0,1)\cup(1,+\infty)$ and applied the substitution $x\mapsto\frac{1}{x}$ in the second integral, in order to have an integral over $(0,1)$ matching the original integral. – 2017-01-08
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If you really love partial fractions you may note that $$I=\int_{0}^{1}\frac{1-x^{3}+x^{6}}{1+x^{4}+x^{8}}\mathrm dx=\frac{1}{4\sqrt{3}}\int_{0}^{1}\frac{1}{x^{2}+\sqrt{3}x+1}\mathrm dx+\frac{1}{4}\int_{0}^{1}\frac{1}{x^{2}+x+1}\mathrm dx $$ $$+\frac{3}{4}\int_{0}^{1}\frac{1}{x^{2}-x+1}\mathrm dx-\frac{1}{4\sqrt{3}}\int_{0}^{1}\frac{1}{x^{2}-\sqrt{3}x+1}\mathrm dx $$ $$=\frac{1}{2\sqrt{3}}\int_{\sqrt{3}}^{2+\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx+\frac{1}{2\sqrt{3}}\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx $$ $$+\frac{\sqrt{3}}{2}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx-\frac{1}{2\sqrt{3}}\int_{-\sqrt{3}}^{2-\sqrt{3}}\frac{1}{x^{2}+1}\mathrm dx $$ so we have $4$ elementary integrals hence $$I=\color{red}{\frac{5\pi}{12\sqrt{3}}}.$$