Prove Fourier Transformation: $\int_{-\infty}^{\infty} \Theta(t) \sin (w_0 t) e^{- i w t} \,dt =- \frac{w_0}{ w^2-w_0^2 +i \text{sgn}(w)0^{+}}$

Question

Prove: $$\begin{align}\int_{-\infty}^{\infty} \Theta(t) \sin (w_0 t) e^{- i w t} \,dt &=- \frac{w_0}{ w^2-w_0^2 +i \text{sgn}(w)0^{+}}\\\\ &=- \frac{w_0}{w^2-w_0^2}+ \frac{i \pi }{2}(\delta(w-w_0)-\delta(w+w_0))\end{align}$$

where $\delta(x)$ is the Dirac function, $\Theta(x)$ is the Heaviside step function, $\text{sgn}(x)$ is the Sign function.

I think that the 2nd equation may need to use Sokhotski–Plemelj theorem, but there is $\text{sgn}(w)$ in the denominator, so I don't know how to use the identity.

Answer 1

Using the convolution theorem, we have

$$\mathscr{F}\{\sin(\omega_0t)\Theta(t)\}=\frac1{2\pi}\underbrace{\mathscr{F}\{\sin(\omega_0t)\}}_{=i\pi(\delta(\omega +\omega_0)-\delta(\omega- \omega_0))}*\underbrace{\mathscr{F}\{\Theta(t)\}}_{=-\frac{i}{\omega}+\pi \delta(\omega)}$$

Therefore, the Fourier Transform of $\sin(\omega_0t)\Theta(t)$ is given by

$$\begin{align} \mathscr{F}\{\sin(\omega_0t)\Theta(t)\}&=\frac1{2\pi}\int_{-\infty}^\infty \left(i\pi(\delta(\omega -\omega'+\omega_0)-\delta(\omega-\omega' -\omega_0)) \right)\,\left(-\frac{i}{\omega'}+\pi \delta(\omega') \right)\,d\omega'\\\\ &=\frac{1/2}{\omega+\omega_0}+i\frac{\pi}{2}\delta(\omega+\omega_0)-\frac{1/2}{\omega-\omega_0}-i \frac{\pi}{2}\delta(\omega-\omega ') \\\\ &=-\frac{\omega_0}{\omega^2-\omega_0^2}+i\frac{\pi}{2} (\delta(\omega+\omega_0)-\delta(\omega-\omega_0)) \end{align}$$

as was to be shown!

---

In distribution, we have

$$\begin{align} \lim_{\epsilon\to 0^+}\int_{-\infty }^\infty \sin(\omega_0 t)\Theta(t)e^{-i(\omega-i\epsilon)t}\,dt&=\lim_{\epsilon\to 0^+}\int_0^\infty \sin(\omega_0 t)e^{-i(\omega-i\epsilon)t}\,dt\\\\ &=\frac1{2i}\lim_{\epsilon\to 0^+}\int_0^\infty (e^{i(\omega_0-\omega-i\epsilon)t}-e^{i(-\omega_0-\omega-i\epsilon)t})\,dt\\\\ &=\frac1{2i}\lim_{\epsilon\to 0^+}\left(\frac{1}{\epsilon+i(\omega-\omega_0)}-\frac{1}{\epsilon+i(\omega+\omega_0)}\right)\\\\ &=-\lim_{\epsilon\to 0^+}\frac{\omega_0}{(\omega^2-\omega_0^2)-i(2\omega)\epsilon -\epsilon^2}\\\\ &=-\frac{\omega_0}{(\omega^2-\omega_0^2)-i\text{sgn}(\omega)0^+} \end{align}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{-\infty}^{\infty}\Theta\pars{t}\sin\pars{\omega_{0}t} \expo{-\ic\omega t}\,\dd t = \int_{-\infty}^{\infty}\ \overbrace{\pars{-\int_{-\infty}^{\infty} {\expo{-\ic\nu t} \over \nu + \ic 0^{+}}\,{\dd\nu \over 2\pi\ic}}} ^{\ds{\Theta\pars{t}}}\ \sin\pars{\omega_{0}t}\expo{-\ic\omega t}\,\dd t \\[5mm] = &\ -\int_{-\infty}^{\infty}{1 \over \nu + \ic 0^{+}} \bracks{% {1 \over 2\ic}\int_{-\infty}^{\infty} \expo{\ic\pars{-\nu + \omega_{0} - \omega}t}\,\dd t - {1 \over 2\ic}\int_{-\infty}^{\infty} \expo{\ic\pars{-\nu - \omega_{0} - \omega}t}\,\dd t}{\dd\nu \over 2\pi\ic} \\[5mm] = &\ -\int_{-\infty}^{\infty}{1 \over \nu + \ic 0^{+}} \bracks{% -\pi\ic\,\delta\pars{-\nu + \omega_{0} - \omega} + \pi\ic\,\delta\pars{-\nu - \omega_{0} - \omega}}{\dd\nu \over 2\pi\ic} \\[5mm] = &\ {1 \over 2}\,{1 \over \omega_{0} - \omega + \ic 0^{+}} - {1 \over 2}\,{1 \over -\omega_{0} - \omega - \ic 0^{+}} \\[5mm] = &\ \bracks{{1 \over 2}\,\mrm{P.V.}{1 \over \omega_{0} - \omega} - {1 \over 2}\,\ic\pi\,\delta\pars{\omega_{0} - \omega}} - \bracks{{1 \over 2}\,\mrm{P.V.}{1 \over -\omega_{0} - \omega} + {1 \over 2}\,\ic\pi\,\delta\pars{-\omega_{0} - \omega}} \\[5mm] = &\ \bbx{-\,{1 \over 2}\,\ic\pi\bracks{\vphantom{\Large A}\delta\pars{\omega + \omega_{0}} - \delta\pars{\omega - \omega_{0}}}}\quad \mbox{because}\quad\mrm{P.V.}{1 \over \pm\omega_{0} - \omega} = 0 \end{align}

See Sokhotski–Plemelj Theorem.