Can anyone simplify the following expression? I guess something from Fourier transform can help:

$$f(\omega) = \lim_\limits{R \to \infty} \frac{1}{R^2} \int_{r=0}^{R}{re^{i \omega r^{-\gamma}}} \mathrm{d}r$$

where $i$ is the imaginary unit, and $\gamma > 2$.

  • $\begingroup$ Look at the individual terms for large $R$. What do we see is happening for say, $c\ge 0$? $\endgroup$ – Mark Viola Jul 15 '15 at 14:57
  • $\begingroup$ So the imaginary unit is not constant? :-) $\endgroup$ – joriki Jul 15 '15 at 15:45
  • $\begingroup$ Please remove the $a(1-(1+bR)e^{-cR})$ part, it is trivial that it has no relevance here. $\endgroup$ – Jack D'Aurizio Jul 15 '15 at 16:46
  • $\begingroup$ If you substantially change the question (in this case removing $a$) when there are already answers, please notify the answers' authors so they can adapt their answers. $\endgroup$ – joriki Jul 16 '15 at 9:59

If you expand the exponential in the integral, only the constant term survives, yielding $R^2/2$, so the limit is $1/2$. This should in fact work for any $\gamma\gt0$, though one might have to argue more carefully in that case why the operations are allowed.

  • $\begingroup$ Something might not be correct here. Just consider the integration over the second term of $e^{tr^{-\gamma}}$. We have $\lim\limits_{R \to \infty} \frac{1}{R^2 (2-\gamma)} \left[ -r^{2-\gamma}\right]_{0^{+}}^{R}$. So, we may not be able to neglect this term, as $R$ grows large as we always would have zero to power of a negative value ($\gamma > 2$). $\endgroup$ – Jeff Jul 16 '15 at 12:51
  • $\begingroup$ @Hossein: What's $t$? In any case, you can separate the integral from $0$ to, say, $2$ as an additive constant and use the expansion only for $r\gt2$. The series of terms you get from the lower limit $2$ converges to a further additive constant, so it's only the terms from the upper limit $R$ that determine the asymptotic behaviour. $\endgroup$ – joriki Jul 16 '15 at 13:01
  • $\begingroup$ The actual problem is that I am trying to find moment generating function (MGF) of a random variable $X$, which will be simplified to $$E \left[ e^{tX} \right] = \lim_\limits{R \to \infty} \exp \left\{\frac{1}{R^2} \int_{r=0}^{R}{re^{i \omega r^{-\gamma}}} \mathrm{d}r \right\}$$. Now, if the limit of exponent tends to 1/2, then $E \left[ e^{tX} \right] = e^{0.5}$. However, from Taylor expansion we know that $E \left[ e^{tX} \right] = 1 + t E\left[X\right] + t^2 E\left[X^2\right]/2 + \cdots $. These two derivations of MGF do not match, implying that something is wrong $\endgroup$ – Jeff Jul 16 '15 at 14:06
  • $\begingroup$ @Hossein: "The limit tends to" makes no sense. The limit is $x$, and the quantity whose limit we take tends to $x$. In any case, I don't know what you mean by that limit; perhaps you could elaborate? $\endgroup$ – joriki Jul 16 '15 at 14:08
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    $\begingroup$ I found the error! Your integration is correct. $\endgroup$ – Jeff Jul 17 '15 at 7:10

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