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  • 积分题05012014

    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
    \[I = \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx}  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \frac{1}{2}\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .\]
    \[I = {\mathop{\rm Im}\nolimits} T.\]
    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
    \[\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx}  = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).\]
     
    \[ \Rightarrow \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx}  =  - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}\]
    \[F'\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx}  - \int_{ - \infty }^\infty  {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) =  - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.\]
     
    \[ \Rightarrow F\left( m \right) =  - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).\]
     
    \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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  • 原文地址:https://www.cnblogs.com/Eufisky/p/3703466.html
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