傅里叶变换求解积分题2
计算
\[\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx.}\]
解：留数理论的一种解答：
注意到
\[\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} .\]
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*}
\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*}
再令
\[I = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .\]
则
\[I = {\mathop{\rm Im}\nolimits} T.\]
即$\displaystyle T$的虚部为$\displaystyle I$.因此，为了计算积分$\displaystyle I$，只需求出积分
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} \]
即可.先求
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} .\]
求得辅助函数
\[\frac{{P\left( z \right)}}{{Q\left( z \right)}}{e^{i\left( {mz} \right)}} = \frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}\]
在上半平面的奇点只有点$\displaystyle \alpha = - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i$（另一个奇点为$\displaystyle \beta = - \frac{1}{2} - \frac{{\sqrt 3 }}{2}i$）.于是我们有
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = 2\pi i \cdot {\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right).\]
由于
\[{\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\frac{{{e^{i\left( {mz} \right)}}}}{{\left( {z - \alpha } \right)\left( {z - \beta } \right)}} = \frac{{{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}}}}{{\sqrt 3 i}}.\]
\[ \Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} - i\sin \frac{m}{2}} \right).\]
同理亦得
\[\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).\]
\[ \Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}\]
故
\[F'\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) = - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.\]
\[ \Rightarrow F\left( m \right) = - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).\]
\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*}
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*}
立得
\[\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{\pi }{{2\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right){\rm{ = }}\frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).\]
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