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

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

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

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