# Fourier transform of $\int^\infty_0 D(\omega) \cos (\omega t - \delta(\omega))\, d\omega$

Let $x_p(t)$ be a superposition of the fourier modes:

$$x_p(t) = \int^\infty_0 D(\omega) \cos[\omega t - \phi(\omega)] \, d\omega$$

I would like to analyse the fourier transform of the function, thus I try to perform a fourier transform on $x_p(t)$, however I obtain something that is no longer a function of frequency $\omega$.

i.e.

$$\mathcal{F}\{x_p(t)\} = \frac{1}{2\pi}\int^{\omega=\infty}_{\omega=0} D(\omega)\int^{t=\infty}_{t=-\infty} e^{-i\omega t}\cos[\omega t - \phi(\omega)]\,dt \,\, d\omega$$

How do I go about performing the fourier transform for $x_p(t)$? Am I having a misconception in my understanding of fourier transformation?

EDIT: Changing the symbol $\delta(\omega)$ to $\phi(\omega)$, to prevent confusion of a phase with a dirac delta function.

• You are using $\omega$ twice for different purposes. $x_p$ is defined using $\omega$ but then you use the same variable for the Fourier frequency. – copper.hat Jun 24 '18 at 16:05
• Assuming $\delta(\omega)$ is meant to represent a Dirac delta function, the quantity $\cos[\omega t - \delta(\omega)]$ is not defined. – John Barber Jun 24 '18 at 18:41
• @JohnBarber Sorry for the confusion, $\delta(\omega)$ is simply a phase that is a function of the frequency $\omega$ – Tian Jun 25 '18 at 2:18
• @copper.hat This problem is actually taken from a dirac delta driving force on a damped driven harmonic oscillator. I first arrived to this expression for $x_p(t)$ by considering individual fourier modes of the dirac delta function, I then take the integral of the frequencies to determine the full expression of $x_p(t)$ – Tian Jun 25 '18 at 2:21
• Perhaps you missed what I wrote. The expression for $x_p$ uses the variable $\omega$, which is fine, but you seem to be reusing the same variable to take the Fourier transform, which is not fine. – copper.hat Jun 25 '18 at 2:54

As noted by @copper.hat in the comments above, your main problem is that you're using the variable $\omega$ twice. The dummy variable of integration in your very first expression should be given a different name, $\omega'$ say: $$x_p(t) \;=\; \int_0^{\infty}d\omega'\, D(\omega')\, \cos[\omega' t - \phi(\omega')]$$ Then: \begin{align*} FT\left\{x_p(t)\right\} &= \frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t}\, x_p(t)\\[0.1in] &= \frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t} \int_0^{\infty}d\omega'\,D(\omega')\, \cos[\omega' t - \phi(\omega')]\\[0.1in] &= \int_0^{\infty}d\omega'\,D(\omega')\,\frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t}\cos[\omega' t - \phi(\omega')]\\[0.1in] &= \int_0^{\infty}d\omega'\,D(\omega')\,\frac{1}{2}\,\frac{1}{2\pi}\int_{-\infty}^{+\infty} dt\, e^{-i \omega t} \left[e^{+i\omega't}e^{-i \phi(\omega')} \,+\,e^{-i\omega't}e^{+i \phi(\omega')}\right]\\[0.1in] &= \frac{1}{2}\int_0^{\infty}d\omega'\,D(\omega')\,\left[\delta(\omega' - \omega)e^{-i \phi(\omega')} \,+\,\delta(\omega'+\omega)e^{+i \phi(\omega')}\right]\\[0.1in] &= \frac{1}{2}\times \begin{cases} D(\omega)\, e^{-i \phi(\omega)} & \omega > 0\\[0.05in] D(-\omega)\, e^{+i \phi(-\omega)} & \omega < 0\\[0.05in] D(0)\,\cos\phi(0) & \omega = 0 \end{cases} \end{align*} In going from the third to the fourth line above, I have used the complex expression for $\cos$: $$\cos x = \frac{e^{+i x} - e^{-i x}}{2}\, .$$ In going from the fourth to the fifth line above, I have used the integral representation of the Dirac delta function: $$\delta (\omega) = \frac{1}{2\pi} \int_{-\infty}^{+\infty}dt\, e^{\pm i \omega t}$$