Which is the transform/result of $\int_a^b \delta(t-a)e^{-iwt}dt$?

Question

Which is the transform/result of $\int_a^b \delta(t-a)e^{-iwt}dt$?

I am trying to analyze some Fourier Transforms for time-limited functions, so the transform will have finite integration limits.

I believe that for the time shift the result will be the same treatment than in the standard Fourier transform, this meaning a phase term $e^{-iaw}$ so: $\int_a^b \delta(t-a)e^{-iwt}dt = e^{-iaw}\int_a^b \delta(t)e^{-iwt}dt$.

But for the $\delta(t)$ distribution I don´t really know how to proceed, since the standard "table" transform $\int_{-\infty}^{\infty} \delta(t)e^{-iwt}dt = 1$ is not longer true.

Thinking in use that $\frac{d\theta(t)}{dt} = \delta(t)$ with $\theta(t)$ the standard unitary step distribution (Heavyside's step function), I think the result is going to be: $$\int_{a}^{b} \delta(t)e^{-iwt}dt = \theta(b)-\theta(a)$$ So the result to my question is going to be: $$\int_{a}^{b} \delta(t-a)e^{-iwt}dt = e^{-iaw}\cdot\left(\theta(b)-\theta(a)\right)$$

But if I try to solve it through Wolfram-Alpha as I did here, many more terms are shown, so I believe I have a misconception (don´t know if the Heavyside's step function derivative property is applicable for these limited integrations limits).

Hope you can help me to figure it out. Beforehand thanks you very much.

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Added later:

Since what I read in the comments, seems some context is needed (and honestly y get lost reading the links you have share with me).

I am trying to understand time-limited functions since a read that analytic functions cannot be described by them, so if $x(t)$ is a standard function and the time limited function is then $f(t) = x(t)\cdot\left(\theta(t-t_f)-\theta(t-t_0)\right)$, with $\theta(t)$ the standard step function, so $f(t)$ is non-zero only between the boundaries $\partial t = \{t_0,\,t_f\}$, then if I try to take its derivative problems will rise at the boundaries because of the discontinuity: $$\frac{df(t)}{dt} = \frac{dx(t)}{dt}\cdot\left(\theta(t-t_f)-\theta(t-t_0)\right)+x(t)\cdot\left(\theta'(t-t_f)-\theta'(t-t_0)\right)$$

I am trying to find upper bounds for the maximum rate of change of this functions when it holds to be finite, so to avoid the problem at the boundaries in the time-domain, I just using: $$\max_t |\frac{df(t)}{dt}| \approx \max_t |\frac{dx(t)}{dt}\cdot\left(\theta(t-t_f)-\theta(t-t_0)\right)|$$

But if I try to move to the frequency domain, the problem get somehow translated to it in the following form: being $\mathbb{F}_{[t_0,\,t_f]}\{f(t)\}(w) = F(w) = \int_{t_0}^{t_f} f(t)\,e^{-iwt}\,dt$ the Fourier Transform of time-limited function, and the Fourier transform of $df(t)/dt = iwF(w)$, every time I take the transform of a function which has some $f(\partial t) \neq 0$ the integral $\int_{-\infty}^\infty |jwF(w)|dw \to \infty$ diverges, and when plotting $|jwF(w)|$ its look like the transform is a function with an added constant.

To avoid this issue I am trying to figure out what I have to subtract to the spectrum, with I believe is going to be equivalent, that being working with the argument of what I am using to estimate the maximum rate of change (because of the linearity of the Fourier Transform): $$\mathbb{F}_{[t_0,\,t_f]}\left\{ \underbrace{\frac{dx(t)}{dt}}_{\text{calculated first}}\cdot\left(\theta(t-t_f)-\theta(t-t_0)\right)\right\} = iw\cdot\mathbb{F}_{[t_0,\,t_f]}\left\{f(t)\right\} -\int\limits_{t_0}^{t_f} x(t)\delta(t-t_f)e^{-iwt}dt + \int\limits_{t_0}^{t_f} x(t)\delta(t-t_0)e^{-iwt}dt$$

But I am stuck because I don't know what to do with the delta functions, its look like the sifting property but because of the finite integration limits I believe it don't apply...

Is $$\int\limits_{t_0}^{t_f} x(t)\delta(t-t_0)e^{-iwt}dt = x(t_0)e^{-iwt_0}$$ ???

I tried for an example and the sign was the opposite I was expecting... I believe that $$\mathbb{F}_{[t_0,\,t_f]}\left\{ \underbrace{\frac{dx(t)}{dt}}_{\text{calculated first}}\cdot\left(\theta(t-t_f)-\theta(t-t_0)\right)\right\} = iw\cdot\mathbb{F}_{[t_0,\,t_f]}\left\{f(t)\right\} -x(t_0)e^{-iwt_0}+x(t_f)e^{-iwt_f}$$ Is that right? I am lost here.

Hope you can help me.

Answer 1

So the short answer is that the notation $$ \int_a^b \delta(t-a)\,e^{-i\omega\,t}\,\mathrm d t $$ does not make sense.

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Here is the longer answer. So first the Dirac delta $\delta_a(t) = \delta(t-a)$ is not a function, it is either defined as a distribution of order $0$ or as a bounded measure. As a measure, it is defined as acting on a set $A$ by $$ \delta_a(A) = \left\{ \begin{array}{} 1 &\text{ if } a∈ A \\ 0 &\text{ if } a\notin A. \end{array}\right. $$ Similarly as for the Lebesgue measure $\mathrm d t$, one can then define the integration with respect to this new measure $\delta_a(\mathrm d t)$ by $$ ∫_{\mathbb{R}} \varphi(t)\, \delta_a(\mathrm{d}t) = \varphi(a) $$ for any nice function $\varphi$. In the same way if $f∈L^1$ is a classical function, it can be identified with the measure $\mu_f(\mathrm{d}t) = f(t)\,\mathrm{d}t$ verifying $$ ∫_a^b \varphi(t)\,\mu_f(\mathrm{d}t) = ∫_a^b \varphi(t)\,f(t)\,\mathrm{d}t = ∫_{[a,b]} \varphi(t)\,f(t)\,\mathrm{d}t = ∫_{(a,b)} \varphi(t)\,f(t)\,\mathrm{d}t $$ Notice however that the last identity is a special feature of locally integrable functions (coming from the fact that the integral is the same if we remove a set of measure $0$) which allows to use the notation $\int_a^b$. This is not the case of the dirac measure. Therefore we have to specify if we are looking at the integral of the dirac over $[a,b]$ or $(a,b)$. And then we have $$ \begin{align*} \int_{[a,b]} \varphi(t)\,\delta_a(\mathrm{d}t) &= \varphi(a) \\ \int_{(a,b)} \varphi(t)\,\delta_a(\mathrm{d}t) &= 0. \end{align*} $$

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So the full answer is $$ \begin{align*} \int_a^b e^{-i\omega\,t}\,\delta_a(\mathrm{d}t) &\text{ is not a clear notation} \\ \int_{[a,b]} e^{-i\omega\,t}\,\delta_a(\mathrm{d}t) &= e^{-i\omega\,a} \\ \int_{(a,b)} e^{-i\omega\,t}\,\delta_a(\mathrm{d}t) &= 0. \end{align*} $$