Prove that,

$$ \int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8} $$

What kind of subsititution should be used to solve this integral

Another integral that give the same answer but with a different limit and same denominator?

$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi}{\phi^5-8} $$

  • $\begingroup$ See what WA gives $\endgroup$ – Archis Welankar Apr 29 '16 at 11:14
  • $\begingroup$ At first glance, I would use complex analysis and a residue theorem, but that doesn't seem to be the intent of this question? $\endgroup$ – Michael Burr Apr 29 '16 at 11:14
  • $\begingroup$ @ArchisWelankar Wolfram gives me $$\int \frac{x^8+1}{x^{10}+1} \, dx=\frac{1}{10} \left(4 \tan ^{-1}(x)+\left(\sqrt{5}-1\right) \tan ^{-1}\left(\frac{\sqrt{2 \left(\sqrt{5}+5\right)}-4 x}{1-\sqrt{5}}\right)+\left(\sqrt{5}-1\right) \tan ^{-1}\left(\frac{4 x+\sqrt{2 \left(\sqrt{5}+5\right)}}{\sqrt{5}-1}\right)+\left(\sqrt{5}+1\right) \tan ^{-1}\left(\frac{4 x-\sqrt{10-2 \sqrt{5}}}{\sqrt{5}+1}\right)+\left(\sqrt{5}+1\right) \tan ^{-1}\left(\frac{4 x+\sqrt{10-2 \sqrt{5}}}{\sqrt{5}+1}\right)\right) $$ but doesn't provide a step-by-step solution. $\endgroup$ – Math1000 Apr 29 '16 at 11:44
  • $\begingroup$ I have came across a particular high-school examination question (for reference: dropbox.com/s/q9yqpags9a1z9m9/…) that asks to evaluate $$\int_{0}^1 \frac{1 + x^4}{1 + x^6}$$ by first proving that for $n \in \mathbb{Z}^+$, $$\frac{1}{1 + x^2} = 1 - x^2 + x^4 - \dots + (-1)^{n - 1}x^{2n - 2} + (-1)^n\frac{x^{2n}}{1 + x^2}$$ ... though I am not sure how applicable it is here. $\endgroup$ – Yiyuan Lee Apr 29 '16 at 12:31
  • $\begingroup$ I must remark that this integral is absolutely beautiful. $\endgroup$ – MathematicsStudent1122 Apr 30 '16 at 6:42

If we make the substitution $t=\frac1x$ and later $x^{10}=u$, then we find that $$\begin{align}\int_0^1\frac{1+x^8}{1+x^{10}}dx&=\int_{\infty}^1\frac{1+t^{-8}}{1+t^{-10}}\frac{(-dt)}{t^2}=\int_1^{\infty}\frac{t^8+1}{t^{10}+1}dt\\ &=\frac12\int_0^{\infty}\frac{1+x^8}{1+x^{10}}dx=\frac12\int_0^{\infty}\frac{1+u^{\frac45}}{1+u}\frac{du}{10u^{\frac9{10}}}=\\ &\frac1{20}\int_0^{\infty}\frac{u^{\frac1{10}-1}+u^{\frac9{10}-1}}{1+u}du\end{align}$$ Now, $$\int_0^{\infty}\frac{u^{p-1}}{1+u}du=\frac{\pi}{\sin p\pi}$$ Is a famous integral that converges for $p\in(0,1)$ and may be determined using the 'keyhole' contour. Of course $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)=\cos\left(\frac{2\pi}5\right)=\frac1{2\phi}$, where $\phi=\frac{\sqrt5+1}2$. So $$\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{2\pi}{20\sin\left(\frac{\pi}5\right)}=\frac{\pi}{\left(\frac5{\phi}\right)}$$ So we just have to compute $$\begin{align}\phi^5-8&=\left(\frac{\sqrt5+1}2\right)^5-8=\frac{11+5\sqrt5}2-8\\ &=\frac{5\sqrt5-5}2=5\left(\frac{\sqrt5-1}2\right)\\ &=5\left(\frac{\sqrt5+1}2\right)^{-1}=\frac5{\phi}\end{align}$$ To conclude $$\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}$$ Now, you can do this with partial fractions, but is seems much more long-winded.


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