Find this integral $$I=\int_{-\pi}^{\pi}\frac{x\cdot \sin(x) \cot^{-1}{(2014^x)}}{1+\cos^4(x)}dx$$

let $x\to -x$,so $$I=\int_{-\pi}^{\pi}\dfrac{x\sin(x) \cot^{-1}{(2014^{-x})}}{1+\cos^4(x)}dx$$ and note $$\cot\cot^{-1}{2014^x}+\cot\cot^{-1}{2014^{-x}}=\dfrac{\pi}{2}$$ so $$2I=\dfrac{\pi}{2}\int_{-\pi}^{\pi}\dfrac{x\sin(x)}{1+(\cos{x})^4}dx=\pi\int_{0}^{\pi}\dfrac{x\sin{x}}{1+(\cos{x})^4}dx=\pi^2\int_{0}^{\pi/2}\dfrac{\sin{x}}{1+(\cos{x})^4}dx$$ so $$I=\dfrac{\pi^2}{2}\int_{0}^{1}\dfrac{1}{1+x^4}dx$$ I feel it's very ugly. Could you please help me using easy methods? It is said that we can use the Gamma function?

  • $\begingroup$ In a pinch, you can always expand the integrand as a geometric series, switch the order of summation and integration, and then re-sum the series. $\endgroup$ – David H Sep 23 '14 at 3:18
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    $\begingroup$ It's not pretty. $\endgroup$ – Chantry Cargill Sep 23 '14 at 3:42
  • $\begingroup$ You can factorise the denominator as $(x+i^{\frac 32})(x - i^{\frac 32})(x + i^{\frac 12})(x - i^{\frac 12})$ then do complex partial fractions. If it was an indefinite integral, putting it back into a nicer form would be pure hell, but since it's a definite integral, it should be evaluable without too much difficulty. I think. EDIT: Mucked around with it a bit. It will still be quite tedious. $\endgroup$ – Deepak Sep 23 '14 at 4:18
  • $\begingroup$ If you're really dead-set on using partial fractions, you could notice that $$\frac{8}{4+u^4}=\frac{2+u}{2+2u+u^2}+\frac{2-u}{2-2u+u^2}$$ and make the appropriate substitutions. karvens' solution is way nicer, though... $\endgroup$ – Micah Sep 23 '14 at 5:43

Substitution by $x+x^{-1}=u$ and $x-x^{-1}=v$ clean things up a bit. \begin{align} 2\int_0^1\frac{1}{1+x^4}\,\mathrm{d}x&=\int_0^1\frac{1+x^{-2}}{x^{-2}+x^2}\,\mathrm{d}x-\int_0^1\frac{1-x^{-2}}{x^{-2}+x^2}\,\mathrm{d}x\\ &=\int_{-\infty}^0\frac{1}{v^2+2}\,\mathrm{d}v+\int_{2}^{\infty}\frac{1}{u^2-2}\,\mathrm{d}u\\ &=\frac{\pi}{2\sqrt{2}}+\frac{\log(\sqrt{2}+1)}{\sqrt{2}} \end{align}


We have \begin{equation} I=\frac{\pi^2}{2}\int_0^1\frac{1}{1+x^4}dx\tag{1} \end{equation} Substituting $t=x^4$ yields \begin{equation} I=\frac{\pi^2}{8}\int_0^1\frac{t^{-\frac{3}{4}}}{1+t}dt \end{equation} According to Part 11. Scientia 18, 2009, 61-75 by Khristo N. Boyadzhiev, Luis A. Medina, and Victor H. Moll, the above integral is defined as the incomplete beta function \begin{equation} \beta(a)=\int_0^1\frac{t^{a-1}}{1+t}dt=\frac{1}{2}\left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}\right) \right] \end{equation} It can easily be proved by multiplying the integrand by $\dfrac{1-t}{1-t}$ and expanding each of the integrands as a geometric series. The details proof can be seen in the cited paper. So \begin{align} I&=\frac{\pi^2}{8}\beta\left(\frac{1}{4}\right)=\frac{\pi^2}{16}\left[\psi\left(\frac{5}{8}\right)-\psi\left(\frac{1}{8}\right) \right]\tag{2} \end{align} Both results in $(1)$ and $(2)$ are numerically agreed to 50 digits precision \begin{equation} I\approx4.2783402057377873840521430671212757030095999944703 \end{equation}


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