<s>[INST] A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$ I need to find a value of this definite integral:
$$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$
Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate. 
Do you have any ideas how I can approach this problem?
 [/INST] There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.
Notice
$$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right]
= \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]
$$
and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have
$$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{2-\cos x}{(1+x^4)(5-4\cos x)} dx =
\frac12\int_{-\infty}^\infty \frac{1}{(1+z^4)(2 - e^{iz})}dz$$
Since $\displaystyle\;\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\Im z \ge 0$ and $\displaystyle\;\left|\frac{1}{2-e^{iz}}\right| \le \frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and
$$\mathcal{I} = \lim_{R\to\infty}\frac12 \oint_{C_R} \frac{1}{(1+z^4)(2 - e^{iz})}dz
\quad\text{ where }\quad
C_R = [-R, R ] \cup \big\{\; Re^{i\theta} : \theta \in [0,\pi]\big\}
$$
$\displaystyle\;\frac{1}{1+z^4}$ has $4$ poles $\omega_k = e^{\frac{(2k+1)i}{4\pi}}, k = 0..3$ over $\mathbb{C}$. Two of them $\omega_0 = \frac{1+i}{\sqrt{2}}$ and $\omega_1 = \frac{-1+i}{\sqrt{2}}$ belongs to the upper half-plane. Since
$$\frac{1}{1+z^4} = \sum_{k=0}^3 \frac{1}{(z-\omega_k) 4\omega_k^3}
= -\frac14 \sum_{k=0}^3\frac{\omega_k}{z-\omega_k}$$
The residue of the integrand at $\omega_k$ are $\displaystyle\;-\frac14 \frac{\omega_k}{2 - e^{i\omega_k}}$ for $k = 0, 1$. This leads to
$$\begin{align}
\mathcal{I} 
&= \frac12\left[ -\frac{2\pi i}{4}\left(\frac{\omega_0}{2 - e^{i\omega_0}} + \frac{\omega_1}{2 - e^{i\omega_1}} \right)\right]\\
&= \frac{-\pi i}{4\sqrt{2}}\left(
\frac{1+i}{2 - e^{-1/\sqrt{2}} e^{i/\sqrt{2}}} +
\frac{-1+i}{2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}}\right)\\
&= \frac{-\pi i}{4\sqrt{2}}
\left[
\frac{
  (1+i)(2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}})
+ (-1+i)(2 -e^{-1/\sqrt{2}} e^{ i/\sqrt{2}})
}{
4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}}
}
\right]\\
&=
\frac{\pi}{2\sqrt{2}}
\left[
\frac{2 - e^{-1/\sqrt{2}}\left(
\cos\left(\frac{1}{\sqrt{2}}\right)
- \sin\left(\frac{1}{\sqrt{2}}\right)
\right)
}{
4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}}
}
\right]\\
&=
\frac{\pi e^{1/\sqrt{2}} }{2\sqrt{2}} 
\left[
\frac{2 e^{1/\sqrt{2}} - \left(
\cos\left(\frac{1}{\sqrt{2}}\right)
- \sin\left(\frac{1}{\sqrt{2}}\right)
\right)
}{
4 e^{\sqrt{2}} - 4 e^{1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + 1
}
\right]
\end{align}
$$
Reproducing what Vladimir get in his answer.
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