How to evaluate integrals where the sifting property of the delta function does not appear to hold?

Question

According to Wolfram MathWorld, the delta function can be defined in terms of a Fourier Transform as $\delta(x) = \mathcal{F}_k[1](x) = \int_{-\infty}^{\infty} e^{-2\pi ikx}dk$, which can be expressed as $\delta(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega t}dt$ using a change of variables.

Also according to Wolfram Mathworld(Sorry for the lack of reference. I was only allowed two. Just google "wolfram fourier transform"), the Fourier Transform and Inverse Fourier Transform can be defined as $$H(\omega) = \mathcal{F}[h(t)] = \int_{-\infty}^{\infty}h(t)e^{-i\omega t}dt \tag{1} \label{1}$$ and $$h(t) = \mathcal{F}^{-1}[H(\omega)] = \frac{1}{2\pi}\int_{-\infty}^{\infty}H(\omega)e^{i\omega t}d\omega \tag{2} \label{2}$$ respectively.

Given a function $H(\omega) = 2\pi\frac{\delta(\omega)}{e^{i\omega}-1}$, how do you find its Inverse Fourier Transform, $\mathcal{F}^{-1}[H(\omega)]$? According to $(\ref{2})$, the Inverse Fourier Transform is given by $$h(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\frac{\delta(\omega)}{e^{i\omega}-1}e^{i\omega t}d\omega. \tag{3} \label{3}$$

Usually with integrals that I have encountered involving the delta function, the sifting property (also described in Wolfram MathWorld) can be used. However, in this case, according to my understanding, the sifting property cannot be used because the function in the integrand multiplying the delta function, namely $\frac{2\pi e^{i\omega t}}{e^{i\omega}-1}$, is not sufficiently smooth at $\omega = 0$ (the limit from the left and the limit from the right are not equal). How then can $(\ref{3})$ be evaluated? Using Wolfram Alpha the answer is shown to be $h(t) = t - \frac{1}{2}$.

Answer 1

I believe I've found an acceptable answer. Starting with the expression for $H(\omega)$, $$H(\omega) = 2\pi\frac{\delta(\omega)}{e^{i\omega}-1},$$ this can be split into real and imaginary parts by first multiplying the numerator and denominator by the complex conjugate of the denominator, namely $$H(\omega) = 2\pi\frac{\delta(\omega)}{e^{i\omega}-1}\frac{e^{-i\omega}-1}{e^{-i\omega}-1} = 2\pi(e^{-i\omega}-1)\frac{\delta(\omega)}{2-2\cos(\omega)},$$ where Euler's formula has been used. Using Euler's formula again we get $$H(\omega) = 2\pi(\cos(\omega)-1 - i\sin(\omega))\frac{\delta(\omega)}{2-2\cos(\omega)}.$$ Distributing yields $$H(\omega) = \pi\delta(\omega)\frac{\cos(\omega)-1}{1-\cos(\omega)} - \pi i\delta(\omega)\frac{\sin(\omega)}{1-\cos(\omega)} = -\pi\delta(\omega)- \pi i\delta(\omega)\frac{\sin(\omega)}{1-\cos(\omega)}.$$ The trick is to note that since $\delta(\omega) = 0$ for all $\omega \ne 0$, then small angle approximations can be used (since the second term will therefore be $0$ except at $\omega = 0$) and, I think, become exact. Using these small angle "approximations" yields $$H(\omega) = -\pi\delta(\omega) - \pi i\delta(\omega)\frac{\omega}{1-(1-\frac{\omega^2}{2})} = -\pi\delta(\omega) - 2\pi i\frac{\delta(\omega)}{\omega}.$$ According to Wolfram MathWorld, $-\frac{\delta(\omega)}{\omega} = \delta'(\omega)$. Using this yields $$H(\omega) = -\pi\delta(\omega) + 2\pi i\delta'(\omega),$$ which can be recognized as the Fourier transform of $$h(t) = t - \frac{1}{2}.$$

Update:

I've been able to solve similar problems by using Taylor or Laurent series expansions for terms multiplied by $\delta(\omega)$. Then since $\delta(\omega)$ is only nonzero where $\omega = 0$, the terms with positive powers of $\omega$ ,i.e. $\omega^1, \omega^2,...$, disappear and the terms with negative powers remain ($\omega^{-1}, \omega^{-2},...$).

Answer 2

Let's consider a distribution of the form

$$f(x)\delta^{(n)}(x-a)$$

Where $f(x)$ is an arbitrary ordinary function. According to the definition of FT we have:

$$\langle\mathcal{F}f(x)\delta^{(n)}(x-a))(k), \phi(k)\rangle = \langle f(x)\delta^{(n)}(x-a), \tilde\phi(x)\rangle$$

Hence by definition we get

$$\langle f(x)\delta^{(n)}(x-a), \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\text{d}k\ e^{-ikx}\tilde\phi(k)\rangle$$

So

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\text{d}k\ \phi(k)\langle f(x)\delta^{(n)}(x-a), e^{-ikx}\rangle$$

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\ \text{d}k\ \phi(k)\langle \delta^{(n)}(x-a), f(x)e^{-ikx}\rangle$$

Hence finally

$$ = (-1)^n\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\ \text{d}k\ \phi(k)\frac{\text{d}^n f(x)e^{-ikx}}{\text{d}x^n}\bigg|_{x = a}$$

So we end upin the definition of a Fourier Transform about the general distribution

$$f(x)\delta^{(n)}(x-a)$$

as

$$\boxed{\mathcal{F}f(x)\delta^{(n)}(x-a))(k) = (-1)^n\frac{1}{\sqrt{2\pi}}\frac{\text{d}^n f(x)e^{-ikx}}{\text{d}x^n}\bigg|_{x = a}}$$

Can you apply this to your function?

In your case $n = 1$ because you have a simple Delta.

$$f(x) = \frac{2\pi\delta(x)}{e^{ix}-1}$$