# 傅里叶变换求解积分题2

Eufisky posted @ 2014年5月01日 04:21 in 数学分析 with tags 傅里叶变换 , 987 阅读

$\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx.}$

$\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} .$

\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*}

$I = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{\sin \left( {mx} \right)}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{sin\left( {mx} \right)}}{{{x^2} - x + 1}}dx} ,T = \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \frac{1}{2}\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} .$
$I = {\mathop{\rm Im}\nolimits} T.$

$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx}$

$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} .$

$\frac{{P\left( z \right)}}{{Q\left( z \right)}}{e^{i\left( {mz} \right)}} = \frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}$

$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = 2\pi i \cdot {\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right).$

${\mathop{\rm Re}\nolimits} s\left( {\frac{{{e^{i\left( {mz} \right)}}}}{{{z^2} + z + 1}}, - \frac{1}{2} + \frac{{\sqrt 3 }}{2}i} \right) = \mathop {\lim }\limits_{z \to \alpha } \left( {z - \alpha } \right)\frac{{{e^{i\left( {mz} \right)}}}}{{\left( {z - \alpha } \right)\left( {z - \beta } \right)}} = \frac{{{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}}}}{{\sqrt 3 i}}.$

$\Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m - \frac{1}{2}im}} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} - i\sin \frac{m}{2}} \right).$

$\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + i\sin \frac{m}{2}} \right).$

$\Rightarrow \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} = - \frac{{4\pi i}}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}$
$F'\left( m \right) = I = {\mathop{\rm Im}\nolimits} T = {\mathop{\rm Im}\nolimits} \frac{1}{2}\left( {\int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} + x + 1}}dx} - \int_{ - \infty }^\infty {\frac{{{e^{i\left( {mx} \right)}}}}{{{x^2} - x + 1}}dx} } \right) = - \frac{{2\pi }}{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \frac{m}{2}.$

$\Rightarrow F\left( m \right) = - \frac{{2\pi }}{{\sqrt 3 }} \cdot \left[ {\frac{{{e^{ - \frac{{\sqrt 3 }}{2}m}}}}{2}\left( { - \cos \frac{m}{2} - \sqrt 3 \sin \frac{m}{2}} \right)} \right] = \frac{\pi }{{\sqrt 3 }} \cdot {e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right).$

\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*}

\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*}

$\int_0^\infty {\frac{{\cos \left( {mx} \right)}}{{{x^4} + {x^2} + 1}}dx} = \frac{\pi }{{2\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\left( {\cos \frac{m}{2} + \sqrt 3 \sin \frac{m}{2}} \right){\rm{ = }}\frac{\pi }{{\sqrt 3 }}{e^{ - \frac{{\sqrt 3 }}{2}m}}\sin \left( {\frac{m}{2} + \frac{\pi }{6}} \right).$

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