I came across the following discussion concerning the computation of the Fourier series of a truncated Gaussian:

Fourier transform of a truncated Gaussian function

Numerical simulations suggest that the decay of such a series depends both on the truncation $T$ and the scale $a$ (see link above for the symbols). For example, for a fixed $T$, one would expect the series to decay more rapidly as $a$ decreases (as is the case when $T=\infty$). This, however, does not seem to be the case. There seems to be some form of "boundary effect" arising from the truncation (which depends on the relation between $a$ and $T$).

Is there a rigorous way of justifying this phenomena?

  • 1
    $\begingroup$ What convergence are you referring to? I would expect the behaviour to be similar to the Fourier transform of the Heaviside function (or its derivative the Dirac distribution). $\endgroup$ Nov 30, 2015 at 10:46
  • $\begingroup$ @Justpassingby I meant "decay", not "convergence". Sorry about the typo. $\endgroup$
    – Rajat
    Nov 30, 2015 at 10:49
  • $\begingroup$ Thanks for the clarification. My expectation remains similar to my previous comment: the (lack of fast) decay of the Fourier transform of a function is related to the presence of steep jumps in the original function; so the decay would resemble that of the Fourier transform of the Heaviside function (not better than 1/t if I remember correctly), albeit multiplied by a constant that decreases linearly with the height of the jump, which height decreases itself exponentially with T or 1/a. $\endgroup$ Nov 30, 2015 at 13:09

1 Answer 1


The Fourier transform of the ordinary product of two functions is the convolution product of their individual Fourier transforms.

The truncated Gaussian is the product of the ordinary Gaussian with the indicator function of the interval $[-T,T].$

The Fourier transform of the ordinary Gaussian is another Gaussian, with variance inversely proportional to the variance of the original. Gaussians are rapidly decreasing functions; see any text on Fourier analysis for an exact definition of 'rapid'.

The Fourier transform of the indicator function of an interval has decay worse than $1/t$ near infinity, i.e., its product with the linear function $t$ does not converge to 0 near infinity.

The convolution of these two has the same bad decay property.


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