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What is the limit of this integral: (Assuming $g(k)$ is smooth and continuous and zero at the end points of integration.)

$$\lim_{y\to\infty}\int g(k) e^{iky}dk = ??$$

The physics that this equation represents suggest that the limit should equal zero. Also using arbitrary test functions for $g(k)$ that satisfy the assumptions results in zero as $y$ becomes large. I have come up with a hand-wavy proof by partially discretizing the integral and assuming that $g(k)$ is approximately constant over $\Delta k$.

$$\lim_{y\to\infty}\int g(k) e^{iky}dk = \lim_{y\to\infty}\sum_j g(k_j)\int_{k_j}^{k_j + \Delta k} (\cos(ky) + i\sin(ky))dk$$

With the further assumption that $\Delta k = 2\pi/y$ each of the integrals in the discretization goes to zero.

It would be nice to have a cleaner and more rigorous way to prove this.

EDIT: I should also mention that $g(k)$ is a complex function, $0 < k$, $0\le y<\infty$, and $k$ & $y$ are both real.

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  • $\begingroup$ This depends on what $g(k)$ is. Obviously this won't be the case if $g(k)=e^{-iky}$. $\endgroup$ Commented Jun 4, 2013 at 18:57
  • $\begingroup$ $g(k)$ is not a function of $y$. I think I may have been over thinking this as I have made some head way using integration by parts. $\endgroup$
    – OSE
    Commented Jun 4, 2013 at 19:06

1 Answer 1

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This is the Riemann–Lebesgue lemma.

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  • $\begingroup$ This is exactly what I needed, thank you! $\endgroup$
    – OSE
    Commented Jun 4, 2013 at 19:55

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