注意到

\int_0^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}  = \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} .<script id="MathJax-Element-2" type="math/tex; mode=display">\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} .</script>

若令

\begin{align*}F\left( m \right) &= \int_{ - \infty }^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}  = \int_{ - \infty }^\infty  {\frac{{\cos \left( {mx} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}dx}\\&= \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\left( {x + 1} \right)\cos \left( {mx} \right)}}{{{x^2} + x + 1}}dx}  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\left( {x - 1} \right)\cos \left( {mx} \right)}}{{{x^2} - x + 1}}dx}.\end{align*}<script id="MathJax-Element-3" type="math/tex; mode=display">\begin{align*}F\left( m \right) &= \int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}dx}\\&= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {x + 1} \right)\cos \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {x - 1} \right)\cos \left( {mx} \right)}}{{{x^2} - x + 1}}dx}.\end{align*}</script>

 

\begin{align*} \Rightarrow F‘\left( m \right) &=  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\left( {{x^2} + x} \right)\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx}  + \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\left( {{x^2} - x} \right)sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx}  \\&= \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx}  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx}\end{align*}<script id="MathJax-Element-4" type="math/tex; mode=display">\begin{align*} \Rightarrow F‘\left( m \right) &= - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {{x^2} + x} \right)\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} + \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\left( {{x^2} - x} \right)sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} \\&= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx}\end{align*}</script>

 

再令

I = \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx}  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .<script id="MathJax-Element-5" type="math/tex; mode=display">I = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .</script>

则

I = {\mathop{\rm Im}\nolimits} T.<script id="MathJax-Element-6" type="math/tex; mode=display">I = {\mathop{\rm Im}\nolimits} T.</script>

即\displaystyle T<script id="MathJax-Element-7" type="math/tex">\displaystyle T</script>的虚部为\displaystyle I<script id="MathJax-Element-8" type="math/tex">\displaystyle I</script>.因此，为了计算积分\displaystyle I<script id="MathJax-Element-9" type="math/tex">\displaystyle I</script>，只需求出积分

\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} <script id="MathJax-Element-10" type="math/tex; mode=display">\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} </script>

即可.先求

\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} .<script id="MathJax-Element-11" type="math/tex; mode=display">\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} .</script>

求得辅助函数

\frac{{P\left( z \right)}}{{Q\left( z \right)}}{e^{i\left( {mz} \right)}} = \frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}<script id="MathJax-Element-12" type="math/tex; mode=display">\frac{{P\left( z \right)}}{{Q\left( z \right)}}{e^{i\left( {mz} \right)}} = \frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}</script>

在上半平面的奇点只有点\displaystyle \alpha  =  - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i<script id="MathJax-Element-13" type="math/tex">\displaystyle \alpha = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i</script>（另一个奇点为\displaystyle \beta  =  - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i<script id="MathJax-Element-14" type="math/tex">\displaystyle \beta = - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i</script>）.于是我们有

\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  = 2\pi i \cdot {\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right).<script id="MathJax-Element-15" type="math/tex; mode=display">\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = 2\pi i \cdot {\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right).</script>

由于

{\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\frac{{{e^{i\left( {mz} \right)}}}}{{\left( {z - \alpha } \right)\left( {z - \beta } \right)}} = \frac{{{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}}}}{{\sqrt 3 i}}.<script id="MathJax-Element-16" type="math/tex; mode=display">{\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\frac{{{e^{i\left( {mz} \right)}}}}{{\left( {z - \alpha } \right)\left( {z - \beta } \right)}} = \frac{{{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}}}}{{\sqrt 3 i}}.</script>

 

\Rightarrow \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} - i\sin \frac{m}{2}} \right).<script id="MathJax-Element-17" type="math/tex; mode=display"> \Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} - i\sin \frac{m}{2}} \right).</script>

同理亦得

\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx}  = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).<script id="MathJax-Element-18" type="math/tex; mode=display">\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).</script>

 

\Rightarrow \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx}  =  - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}<script id="MathJax-Element-19" type="math/tex; mode=display"> \Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}</script>

故

F‘\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) =  - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.<script id="MathJax-Element-20" type="math/tex; mode=display">F‘\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) = - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.</script>

 

\Rightarrow F\left( m \right) =  - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).<script id="MathJax-Element-21" type="math/tex; mode=display"> \Rightarrow F\left( m \right) = - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).</script>

 

\begin{align*} \Rightarrow \int_0^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}  &= \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}  = \frac{1}{2}F\left( m \right) \\&= \frac{\pi }{{2\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) = \frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).\end{align*}<script id="MathJax-Element-22" type="math/tex; mode=display">\begin{align*} \Rightarrow \int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} &= \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}F\left( m \right) \\&= \frac{\pi }{{2\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) = \frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).\end{align*}</script>

 

另解：由Fourier变换公式，我们有

\begin{align*}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) &= \frac{2}{\pi }\int_0^\infty  {\cos \left( {mx} \right)dx} \int_0^\infty  {{e^{ - \frac{{\sqrt 3 }}{2}u}}\left( {\cos \frac{u}{2} + \sqrt 3 \sin \frac{u}{2}} \right)\cos \left( {ux} \right)du}  \\&= \frac{{2\sqrt 3 }}{\pi }\int_0^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}.\end{align*}<script id="MathJax-Element-23" type="math/tex; mode=display">\begin{align*}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right) &= \frac{2}{\pi }\int_0^\infty {\cos \left( {mx} \right)dx} \int_0^\infty {{e^{ - \frac{{\sqrt 3 }}{2}u}}\left( {\cos \frac{u}{2} + \sqrt 3 \sin \frac{u}{2}} \right)\cos \left( {ux} \right)du} \\&= \frac{{2\sqrt 3 }}{\pi }\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}.\end{align*}</script>

立得

\int_0^\infty  {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx}  = \frac{\pi }{{2\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right){\rm{ = }}\frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).<script id="MathJax-Element-24" type="math/tex; mode=display">\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{\pi }{{2\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right){\rm{ = }}\frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).</script>