Given a uniform random variable $x$, I want to compute following expectations: $E[e^{\pm j2\pi fx}]$, $E[e^{\pm j2\pi kx}]$, and $E[e^{\pm j2\pi (f+k)x}]$, where $j=\sqrt{-1}$, $k$ is an integer and $f$ is a rational number. I consider two cases: $x\sim\mathcal{U}[0,a]$ and $x\sim\mathcal{U}[-a/2,a/2]$. I tried computing these expectations and arrived at the following answers.
- $E[e^{\pm j2\pi fx}]$: For the case $x\sim\mathcal{U}[0,a]$, $$E[e^{\pm j2\pi fx}] = \int\limits_{0}^{a}e^{\pm j2\pi f x}\frac{1}{a}dx = \frac{1}{\pm j2\pi af}e^{\pm j2\pi f x} |_{0}^{a} = e^{\pm j2\pi f\frac{a}{2}}\frac{1}{\pi af}\frac{(e^{\pm j2\pi f \frac{a}{2}} - e^{\mp j2\pi f \frac{a}{2}}) }{\pm 2j} = e^{\pm j2\pi f\frac{a}{2}}\frac{\sin(\pi af)}{\pi af} = e^{\pm j2\pi f\frac{a}{2}}\text{sinc}(af),$$ where $\text{sinc}(af) = \frac{\sin(\pi af)}{\pi af}$.
When $x\sim\mathcal{U}[-a/2,a/2]$, repeating the above computations, I get $$E[e^{\pm j2\pi fx}] = \int\limits_{-a/2}^{a/2}e^{\pm j2\pi f x}\frac{1}{a}dx = \text{sinc}(af).$$
$E[e^{\pm j2\pi kx}]$: For both distributions, by the definition of Kronecker delta function, I have $$E[e^{\pm j2\pi fx}] = \delta_k.$$
$E[e^{\pm j2\pi (f+k)x}]$: Should this be similar to the first expectation? That is, for $x\sim\mathcal{U}[0, a]$, $$E[e^{\pm j2\pi (f+k)x}] \stackrel{?}{=} e^{\pm j2\pi (f+k)\frac{a}{2}}\text{sinc}(a(f+k)),$$ and, for $x\sim\mathcal{U}[-a/2,a/2]$, $$E[e^{\pm j2\pi (f+k)x}] \stackrel{?}{=} \text{sinc}(a(f+k)).$$
There is a confusion in 2. You claim that the result is the Kronecker delta which is defined only in the discrete case as (note the $k$ in the exponent, there is some mismatch with $f$ in your text):
$E[e^{\pm j2\pi k x}] = \delta_k = {\Large\{ \,} {1 \quad \mathrm{if}\; k=0 \atop 0 \quad \mathrm{else}\quad}$
The "expectation" $E[\; . \; ]$ used in this expression, however, is not the integral as in 1., but in order for the result ($\delta_k$) to be true, $E[\; . \; ]$ must be interpreted as a discrete sum over events, which only works for $1/x = N \in \cal Z$, namely
$E[\; h(k) \; ] = \frac1N \sum_{k=0}^{N-1} h(k) $
In any case, this discrete expectation is not compatible to the other expectations you used, so let's leave this aside. What you really want is not the Kronecker delta but the delta-distribution which is defined as (note there is no $k$ in the exponent):
$E_{\infty}[e^{\pm j2\pi f x}] = \int_{-\infty}^{\infty} e^{\pm j2\pi f x} \rm d x =\delta(f) = {\Large\{ \,} {\infty \quad \mathrm{if}\; f=0 \atop 0 \quad \mathrm{else}\quad}$
where the last "numerical" result is of course to be understood in the distribution sense.
Now this can be retraced from your previous results as follows:
You already have shown
$ E_1 (a) = E[e^{\pm j2\pi fx}] = \int\limits_{-a/2}^{a/2}e^{\pm j2\pi f x}\frac{1}{a}dx = \text{sinc}(af). $
Now use
$$ E_1(a \to \infty) = \lim_{a \to \infty} \int\limits_{-a/2}^{a/2}e^{\pm j2\pi f x}\frac{1}{a}dx = \lim_{a \to \infty} \frac{\delta(f)}{a} \quad {\rm{where}} \; {\rm{ here^* }} \; \delta(f) = {\Large\{ \,} { a \quad \mathrm{if}\; f=0 \atop 0 \quad \mathrm{else}\quad} $$
*: this definition must be made since the limit $a \to \infty $ hasn't been taken yet at this point.
So comparing this with your result $ E_1(a \to \infty) = \lim_{a \to \infty} \text{sinc}(af)$, the two formulations are equal. It is clear that the use of delta-distributions lacks rigor here.
For 3. : your statements are fine. The derivation is as in 1., replacing $f$ with $f+k$. Note that in 1., no assumptions have been made for $f$. Further, following the discussion of 2., you can use this result for any $a$. In particular, as $a \to \infty$, again we have symbolically that $ \text{sinc}(a(f+k)) = \delta(f+k) / a$ which could very unusually be written, as $a \to \infty$, as a Kronecker delta: $ \delta(f+k) / a = \delta_{f+k}$ .