I have a function $g(x)$ defined numerically that is sort of in between a Gaussian and a Lorentzian. It decays much slower than a Gaussian, but still faster than a simple inverse power.

enter image description here

I need to calculate its Fourier transform $f(t)\equiv \mathcal{F}[g(x)](t)$ for large $t$. Because function calls to $g(x)$ are computationally expensive, I define an interpolation of $g(x)$ - call it $g_{\text{int}}(x)$ - on some huge range of $x$, $-40<x<40$, and use that for my integral.

$$f(t)=\int_{-\infty}^{\infty}\cos (tx)g(x)\,dx\,\,\underset{\approx}{\longrightarrow}\,\,\,\int_{-L}^{L}\cos(tx)g_{\text{int}}(x)\,dx$$

However, when I calculate an approximation to the Fourier transform, I get some odd oscillations that I would not initially expect.

enter image description here

As I have indicated in the picture above, the oscillations have a "period" of about 15.7. My first guess would be that this might be an artifact of the alternating nature of cancellation of the integral, but that would not explain the observed "period" of 15.7.

$$T_{\text{guess}}=\frac{2\pi}{L}\approx 0.157\ldots$$

which is exactly a factor of 100 different from what I observe (yes, I have checked that I defined my integrals and horizontal axes correctly). How could this be?

Edit #1: Interpolation Details

I am interpolating with Mathematica's built-in Interpolation, which interpolates between successive points with a cubic curve (so at each point up to the 2$^{\text{nd}}$ derivative is defined). I am specifically interpolating the function $g(x)$ over the range $-40<x<40$ in steps of $dx=40/100=0.4$.

In fact, now that I write that, I realize it could very well be an artifact of my finite sampling, because:

$$T_{\text{guess #2}}=\frac{2\pi}{dx}=\frac{2\pi}{0.4}=15.7\ldots$$

I would appreciate any further help on this, in particular a good way to overcome this problem.

Edit #2: The function $g(x)$

h[x_?NumericQ, En_?NumericQ, pz_?NumericQ] := 
 1./(En^2 + pz^2 + 0.24^2)*
      0.316/(1. + 
         1.2*((k4 + 0.5*En)^2 + kp + (x*pz)^2))^1.*0.316/(1. + 
         1.2*((k4 - 0.5*En)^2 + kp + ((1. - x)*pz)^2))^1.])*((1. - 
         x)*0.316/(1. + 1.2*((k4 + 0.5*En)^2 + kp + (x*pz)^2))^1. + 
      x*0.316/(1. + 
         1.2*((k4 - 0.5*En)^2 + kp + ((1. - x)*pz)^2))^1.))/(((k4 + 
        0.5*En)^2 + 
      kp + (x*pz)^2 + (0.316/(1. + 
         1.2*((k4 + 0.5*En)^2 + kp + (x*pz)^2))^1.)^2)*((k4 - 
        0.5*En)^2 + 
      kp + ((1. - x)*
        pz)^2 + (0.316/(1. + 
         1.2*((k4 - 0.5*En)^2 + 
            kp + ((1. - x)*
              pz)^2))^1.)^2)), {k4, -\[Infinity], \[Infinity]}, {kp, 
    0, \[Infinity]}, Method -> "LocalAdaptive", 
   MaxRecursion -> 
    100]; (*LocalAdaptive seems to work slightly faster *)

g[x_]:=h[0.5,x,2.]; (*this is the function*)
  • $\begingroup$ What kind of interpolation are you using? It is actually not the easiest function to interpolate accurately on such a large domain. $\endgroup$
    – Anton Menshov
    Jul 7, 2018 at 23:06
  • $\begingroup$ @AntonMenshov See my most recent edit. $\endgroup$ Jul 7, 2018 at 23:16
  • 4
    $\begingroup$ I want to point out that every numerical integration method essentially constructs an interpolant of your function and integrates that exactly, so by interpolating the function yourself first, you are in effect replacing Integrate's adaptive construction of interpolants (and all the thought that went into its design) with your own, which is probably a bad idea. $\endgroup$
    – Kirill
    Jul 8, 2018 at 8:42
  • 2
    $\begingroup$ Another idea: given how nice $g$ is, try constructing a Chebyshev series (a kind of well-behaved high-order polynomial interpolation) of $\log g(x)$ on $[-L,L]$ (like so: Module[{n=32,L=10},Exp[InterpolatingPolynomial[Table[{x,Log[g[x]]},{x,-L Cos[(π N@Range[1,2n-1,2])/(2n)]}],x]]], not sure how many nodes you'll need), and then integrating that. (This is not a good way of computing the Chebyshev approximation, just for demonstration.) $\endgroup$
    – Kirill
    Jul 8, 2018 at 21:14
  • 1
    $\begingroup$ I also think that it would help to compile the integrand (gist.github.com/ikirill/931dbf6fb3581f7d66353b74c4743cd3), independent of all the harder issues here, it speeds it up by a factor of ~8 for me. $\endgroup$
    – Kirill
    Jul 9, 2018 at 12:24

1 Answer 1


I want to mention another idea in case it helps. The truncation error of replacing $\int_{-\infty}^{\infty}$ with $\int_{-L}^L$ is on the order of $$\begin{aligned} \int_L^\infty \cos(tx)g(x)\,\mathrm{d}x &= \frac{1}{t}\sin(tx)g(x)\big|_{L}^{\infty} - \int_L^\infty\frac{\sin tx}{t}g'(x)\,\mathrm{d}x \\&= -\frac1L g(L)\sin(tL) + \text{asymptotically smaller terms}. \end{aligned} $$ I took one omitted interval, used integration by parts, and assumed that $g'$ decays faster than $g$, which is usually reasonable.

If $g(L)$ decays like a power and not exponentially or like a Gaussian, then $g(L)$ might not decay fast enough to suppress the leading term in the truncation error of the Fourier integral, and you would then see an oscillating error. One important consequence would be that this error would remain even if the interpolation step is perfectly accurate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.