# Simplifying an expression with Fourier transform

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 $\gamma > 2$.

• Look at the individual terms for large $R$. What do we see is happening for say, $c\ge 0$? – Mark Viola Jul 15 '15 at 14:57
• So the imaginary unit is not constant? :-) – joriki Jul 15 '15 at 15:45
• Please remove the $a(1-(1+bR)e^{-cR})$ part, it is trivial that it has no relevance here. – Jack D'Aurizio Jul 15 '15 at 16:46
• 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. – 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.
• 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 ($\gamma > 2$). – Jeff Jul 16 '15 at 12:51
• @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. – joriki Jul 16 '15 at 13:01
• 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 – Jeff Jul 16 '15 at 14:06
• @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? – joriki Jul 16 '15 at 14:08