$e^{i x} θ(x)$的Fourier transform

hbghlyj

wolframalpha.com/input?i=integral_%28-%E2%88% … %28-i+%CF%89+x%29+dx

$\displaystyle\int_{-\infty}^\infty e^{i x} θ(x)\frac1i e^{-i ω x}\rmd x=\frac1{1-\omega}$ for Im(ω)<0
即
$\displaystyle\int_0^\infty e^{i x}\frac1i e^{-i ω x}\rmd x=\frac1{1-\omega}$ for Im(ω)<0
如何計算

hbghlyj

$\displaystyle\int_{-∞}^∞ e^{i x} \operatorname{sgn}(x)\frac1{2 i} e^{-i ω x}\rmd x = \frac1{1 - ω}$和上面一樣，竟然

hbghlyj

hbghlyj 发表于 2024-2-13 20:41
$\displaystyle\int_0^\infty e^{i x}\frac1i e^{-i ω x}\rmd x=\frac1{1-\omega}$ for Im(ω)<0
如何計算

懂了！For Im(ω)<0，當$x\to+\infty$時$e^{i(1-ω)x}\to0$