Eufisky - The lost book

## 傅里叶变换求解积分题2

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

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

\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*}

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

## 傅里叶变换求积分函数

\begin{align*}&\int_0^\infty  {\frac{{\cos tx}}{{1 + {t^2}}}} dt;\\&\int_0^\infty  {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du} .\end{align*}

\begin{align*}{e^{ - x}} &= \frac{2}{\pi }\int_0^\infty  {cos\left( {\lambda x} \right)d\lambda \int_0^\infty  {{e^{ - u}}} cos\left( {\lambda u} \right)} du = \frac{2}{\pi }\int_0^\infty  {\frac{{\cos \lambda x}}{{{\lambda ^2} + 1}}d\lambda } ;\\{e^{ - x}}\cos x &= \frac{2}{\pi }\int_0^\infty  {\cos \left( {xu} \right)du} \int_0^\infty  {{e^{ - t}}\cos t\cos \left( {ut} \right)dt}  = \frac{2}{\pi }\int_0^\infty  {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du}.\end{align*}

\begin{align*}\int_0^\infty  {\frac{{\cos \lambda x}}{{{\lambda ^2} + 1}}d\lambda }  &= \frac{\pi }{2}{e^{ - x}};\\\int_0^\infty  {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du}  &= \frac{\pi }{2}{e^{ - x}}\cos x.\end{align*}

\begin{align*}\int_0^\infty  {\frac{{\cos xu}}{{{\beta ^2} + {u^2}}}du}  &= \frac{\pi }{{2\beta }}{e^{ - \beta x}} \qquad x > 0,\beta  > 0;\\\int_0^\infty  {\frac{{u\sin xu}}{{{\beta ^2} + {u^2}}}du}  &= \frac{\pi }{2}{e^{ - \beta x}}\qquad x > 0,\beta  > 0;\\\int_0^\infty  {\frac{{{x^{\mu  - 1}}\sin \left( {ax} \right)}}{{{x^2} + 1}}dx}  &=  - {a^{2 - \mu }}\mathbf{\Gamma} \left( {\mu  - 2} \right){}_1{F_2}\left( {1;\frac{{3 - \mu }}{2},\frac{{4 - \mu }}{2};\frac{{{a^2}}}{4}} \right)\mathrm{sign}\left( a \right)\sin \frac{{\mu \pi }}{2}\\&+ \frac{\pi }{2}\sec \frac{{\mu \pi }}{2}\sinh \left( a \right)\qquad{\mathop{\rm Im}\nolimits} a = 0, - 1 < \mathrm{Re}\mu  < 3;\\\int_0^\infty  {\frac{{{x^{\mu  - 1}}\cos \left( {ax} \right)}}{{{x^2} + 1}}dx}  &= \frac{\pi }{{2\sin \frac{{\mu \pi }}{2}}}\cosh a + \frac{1}{2}\cos \frac{{\mu \pi }}{2}\mathbf{\Gamma} \left( \mu  \right)\\&\left[ {{e^{ - a + i\pi \left( {1 - \mu } \right)}}\gamma \left( {1 - \mu , - a} \right) - {e^a}\gamma \left( {1 - \mu ,a} \right)} \right]\qquad a > 0,0 < \mathrm{Re}\mu  < 3;\\\int_0^\infty  {\frac{{{x^{2\mu  + 1}}\sin \left( {ax} \right)}}{{{x^2} + {b^2}}}dx}  &=  - \frac{\pi }{{2\cos \left( {\mu \pi } \right)}}{b^{2\mu }}\mathrm{sinh}\left( {ab} \right) + \frac{{\sin \left( {\mu \pi } \right)}}{{2{a^{2\mu }}}}\mathbf{\Gamma} \left( {2\mu } \right)\\&\left[ {{}_1{F_1}\left( {1;1 - 2\mu ;ab} \right) + {}_1{F_1}\left( {1;1 - 2\mu ; - ab} \right)} \right] \qquad a > 0, - \frac{3}{2} < \mathrm{Re}\mu  < \frac{1}{2};\\\int_0^\infty  {\frac{{{x^{2\mu  + 1}}\cos \left( {ax} \right)}}{{{x^2} + {b^2}}}dx}  &=  - \frac{\pi }{{2\sin \left[ {\left( {\mu  + \frac{1}{2}} \right)\pi } \right]}}{b^{2\mu  + 1}}\cosh \left( {ab} \right) + \frac{{\cos \left[ {\left( {\mu  + \frac{1}{2}} \right)\pi } \right]}}{{2{a^{2\mu  + 1}}}}\mathbf{\Gamma} \left( {2\mu  + 1} \right)\\&\left[ {{}_1{F_1}\left( {1; - 2\mu ;ab} \right) + {}_1{F_1}\left( {1; - 2\mu ; - ab} \right)} \right] \qquad a > 0, - 1 < \mathrm{Re}\mu  < \frac{1}{2}.\end{align*}