(Quick note; I'm a physicist so it means I might have crowbarred some maths into an equation that shouldn't be there, resulting in the problems I'm having.)

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}

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.

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

  • $\begingroup$ Why did the last integral was concluded to be divergent? $\endgroup$ – OR. Jun 16 '17 at 10:37
  • $\begingroup$ Because we are left with an odd function, sin(y), multiplied by another odd function, y, so the integral is of an even function overall. Integrated between zero and infinity this must be divergent... I think $\endgroup$ – Tbone Willsone Jun 16 '17 at 10:41
  • $\begingroup$ Well, I don't know which odd functions but just mind the fact that one can have divergent things which sum is not divergent. $\endgroup$ – OR. Jun 16 '17 at 10:44
  • $\begingroup$ That's true, and it's what I think might be occurring. Are you aware of any way to extract convergent behaviour from a series in which each individual term is divergent? $\endgroup$ – Tbone Willsone Jun 16 '17 at 10:45
  • $\begingroup$ There is whole parade of methods you can start here. $\endgroup$ – OR. Jun 16 '17 at 10:48

Your series has non meaning as the limits of the sin and cos functions, for the argument running to infinity, do not exist. Things are quite different if this is seen in the sense of distributions. Firstly, we note the obvious fact that $$ \int_{-\infty}^\infty\frac{\sin kx}{x}dx= \int_{-\infty}^\infty\frac{\sin x}{x}dx=\pi. $$ Then, you can just Taylor expand the test function and write $$ \int_{-\infty}^\infty\frac{\sin kx}{x}f(x)dx= \pi f(0)+\int_{-\infty}^\infty\frac{\sin kx}{x}xf'(0)dx +\frac{1}{2}\int_{-\infty}^\infty\frac{\sin kx}{x}x^2f''(0)dx+\ldots. $$ That integrals can only be seen as distributions but $$ \int_{-\infty}^\infty\sin kxdx=-i\pi[\delta(k)-\delta(-k)]=0 $$ as the Dirac distribution is even. Then, $$ \int_{-\infty}^\infty x\sin kxdx=\frac{1}{2i}\int_{-\infty}^\infty x[e^{ikx}-e^{-ikx}]=-2\pi\frac{d}{dk}\delta(k) $$ and $$ \int_{-\infty}^\infty x^2\sin kxdx=\frac{1}{2i}\int_{-\infty}^\infty x^2[e^{ikx}-e^{-ikx}]=i\pi\frac{d^2}{dk^2}[\delta(k)-\delta(-k)]=0 $$ and so on, noting that only even terms of the Taylor expansion survive as distributions. In the end you will get a distribution series in $k$ with all the terms, with the exception of the leading one, that will go to zero when $k$ runs to infinity. This limit can only be attached a meaning in the sense of distributions.

  • $\begingroup$ Thank you for the answer :) I actually ended up successfully finding an answer to this this through a Laplace transformation of a convolution integral that my function was involved in. As you say this only made sense in the sense of distributions! $\endgroup$ – Tbone Willsone Jul 20 '17 at 13:36

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