傅里叶变换求积分函数
来自刚哥的虐心的积分题:
\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*}![](/images/smiley/chito/15.gif)
![](/images/smiley/chito/15.gif)
解:事实上,由Fourier变换公式
\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*}
我们得到
\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*}
一般地,找到以下结论
\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*}
2022年8月17日 23:01
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