Calculate $\int_{0}^{\infty}\frac{\cos(x)}{x^2+x+1}dx $ How to calculate integral
$$\int_{0}^{\infty}\frac{\cos(x)}{x^2+x+1}dx $$
It's rather easy to find when the limits of integration are $[-\infty, \infty],$
but with lower limit being zero?
Is it at all computable with some basic knowledge of complex analysis?
Thanks in advance.
 A: By a useful property of the Laplace transform we have
$$ I=\int_{0}^{+\infty}\frac{\cos(x)}{1+x+x^2}\,dx = \frac{2}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sin\left(\frac{\sqrt{3}}{2}s\right)se^{-s/2}}{1+s^2}\,ds \tag{1}$$
with
$$ \int_{0}^{+\infty}\frac{e^{-\alpha s}}{1\pm i s}\,ds = \frac{1}{2}e^{\mp i\alpha}\left(\pi\pm 2i\text{Ci}(\alpha)-2\text{Si}(\alpha)\right)\tag{2}$$
for any $\alpha:\text{Re}(\alpha)>0$. So the original integral depends on sine and cosine integrals.
Its numerical evaluation is not that hard, however: it is enough to apply a few steps of integration by parts, then the Cauchy-Schwarz inequality, to the RHS of $(1)$. This leads to $I\approx\frac{\pi}{7}$, for instance.
A: As Robert Israel answered.
Consider the general problem of $$I=\int \frac {\cos(x)}{(x-a)(x-b)} \,dx$$ and use partial fraction decomposition $$\frac {1}{(x-a)(x-b)}=\frac 1{a-b}\left(\frac 1{x-a} -\frac 1{x-b}\right)$$ So, basically we are left with  integrals
$$J= \int \frac {\cos(x)}{x-c} \,dx=\int \frac {\cos(y+c)}{y} \,dy=\cos (c)\int\frac{ \cos (y)}{y}\,dy-\sin (c)\int\frac{ \sin (y)}{y}\,dy$$that is to say $$J=\cos (c)\, \text{Ci}(y)-\sin (c)\,\text{Si}(y)$$ where appear the sine and cosine integrals. For $y$, the bounds are now $c$ and $\infty$ making $$K=\int_c^\infty \frac {\cos(y+c)}{y} \,dy=\text{Si}(c)\, \sin (c)-\text{Ci}(c) \,\cos (c)-\frac{\pi }{2}  \sin (c)$$ Now, it is sure that with $a=-\frac{1-i \sqrt{3}}{2} $ and $b=-\frac{1+i \sqrt{3}}{2}$, we shall arrive to quite ugly complex expressions for the posted integral.
Its decimal representation is $I\approx 0.451312142585023$ for which $RIES$ proposes things like $\frac{4}{5 \sqrt{\pi }}$ and $\log \left(\frac{1}{3} 2^{2 \phi -1}\right)$; however, $ISC$ did not find anything like this number.
A: Use partial fractions and the Ei function.  It can be converted to incomplete Gamma functions, but
 I don't think the answer can be expressed using only elementary functions. 
