Recovering the delta function with sin⁡(nx)/x

[Tbonewillsone]

Homework Statement

Ultimately, I would like a expression that is the result of an integral with the sin(nx)/x function, with extra terms from the expansion. This expression would then reconstruct the delta function behaviour as n goes to infty, with the extra terms decaying to zero. I understand this might not be possible, and if so I'd like to know why!

Homework Equations

If we have

\begin{equation}
I= \int^{\infty}_{-\infty} \lim_{n \rightarrow \infty}\left( \frac{\sin (n x)}{x} \right) f(x) \mathrm dx=\lim_{n\rightarrow \infty}I_{n},
\end{equation}

then through a change of variables, nx=y

\begin{equation}
I_{n} = \int^{\infty}_{-\infty} \left( \frac{\sin (n x)}{x} \right) f(x) \mathrm dx = \int^{\infty}_{-\infty} \left( \frac{\sin (y)}{y} \right) f \left(\frac{y}{n} \right) \mathrm dy,
\end{equation}

and so

\begin{equation}
I = \int^{\infty}_{-\infty}\left( \frac{\sin (y)}{y} \right) f \left( 0 \right) \mathrm dy = \pi f \left( 0 \right) .
\end{equation}

This replicates the Dirac delta function, meaning that at this limit we can say

\begin{equation}
\lim_{n \rightarrow \infty} \left( \frac{\sin (n x)}{x} \right) \rightarrow \pi \delta(x).
\end{equation}

The Attempt at a Solution

I would like to expand the test function, and then, through taking the limit, recover the delta function property of our function.

\begin{equation}
I_{n} = \int^{\infty}_{-\infty} \left( \frac{\sin (n x)}{x} \right) f(x) \mathrm dx = \int^{\infty}_{-\infty} \left( \frac{\sin (y)}{y} \right) f \left(\frac{y}{n} \right) \mathrm dy,
\end{equation}

Putting the function through a Taylor expansion,

\begin{equation}
\int^{\infty}_{-\infty} \left( \frac{\sin (y)}{y} \right) f \left(\frac{y}{n} \right) \mathrm dy= \int^{\infty}_{-\infty} \left( \frac{\sin (y)}{y} \right) \left( f \left(0\right) + \frac{y}{n}f^{\prime} \left(0\right) + \frac{y^{2}}{n^{2}}f^{\prime \prime} \left( 0 \right) + \mathcal O\left( \frac{y^{3}}{n^{3}} \right) \right) \mathrm dy,
\end{equation}

This integral clearly diverges, I could take the limit at this stage, but I don't understand why this expansion would not work, it must be somehow linked to how "integrateable" the sin(nx)/x is. Is there any way I can manipulate the taylor series to get some converging terms together?

 

[SqueeSpleen]

Why does the integral "clearly diverges"?
Sorry I'm not to used to working with integrals over infinite intervals, but I have a few things to point out, I hope it can help you. The one thing I know is that while you can think of delta function as a weak limit, one natural way to define it is as a continuous functional in the dual space of $H^{1} ( \Omega )$ so you have extra hyphotesis about $f(x)$ that you might use, about integrability of $f$ and it's first derivative (altought I don't think it's the most general way to define it, as it's enough to ask functions to be continuous). But infinite intervals have to be taken with extra care as the integrals in them are not well suited for lebesgue integration (well, improper riemann integration is taking a limit, so it's not really the classic definition with upper and lower sums).

Anyway, one of the things about $sin(x)/x$ is that it's not absolutely integrable, only integrable and this is not usually a good thing, specially when you try to interchange limits for things like taylor series.
If you want to manipulate the series you might want to use the fact that $y/n$ are pair functions to delete those terms from the expression. It's still not enough.

As I was saying, as $sin(x)/x$ is not absolutely integrable, so you might want to integrate it using the principal value to see if it's more well behaved.
$$
\lim_{R \to \infty} \int_{-R}^{R} \dfrac{sin(nx)}{x} dx
$$

You surely can recover delta function if you take a "oblique limit" I mean, make $n$ and $R$ go to infinity together with some special relationship, but I'm sure it would be hard to justify rigorously what you're doing.