Can anyone simplify the following expression? I guess something from Fourier transform can help:
$f(\omega) = \lim_\limits{R \to \infty} \frac{1}{R^2} \int_{r=0}^{R}{re^{ \omega r^{-\gamma}}} \mathrm{d}r$, where $\gamma > 2$.
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
assuming $c>0$, this limit is dominated by the upper end of the integration interval, so $f(\omega)=a/2$ independent of $\omega$; as a check, try $\gamma=3$, when a closed form expression exists,
$$f(\omega) = \lim_\limits{R \to \infty} \frac{a}{R^2}\left( 1 - (1+bR)e^{-cR}\right) \int_{r=0}^{R}{re^{i \omega r^{-3}}} \mathrm{d}r$$ $$=\lim_\limits{R \to \infty}\frac{1}{3}a \left(1-(b R+1) e^{-c R}\right) \Gamma \left(-\frac{2}{3},-\frac{i \omega}{R^3}\right) \left(\frac{i R^3}{\omega}\right)^{-2/3}$$ $$=\lim_\limits{R \to \infty}\frac{1}{3}a \left(1-(b R+1) e^{-c R}\right)\times\frac{3}{2}=\frac{1}{2}a$$
hmm, I notice the question has been heavily edited; in reference to the new question:
$$ \lim_\limits{R \to \infty} \frac{1}{R^2} \int_{0}^{R}{re^{ \omega r^{-\gamma}}} \mathrm{d}r=\frac{1}{2}\;\;{\rm for}\;\;{\rm Re}\,\omega\leq 0.$$
for ${\rm Re}\,\omega>0$ the integral is undefined.
answered Jul 15, 2015 at 17:39
-
$\begingroup$ There is a contradiction with this proof. I posed a question to highlight this contradiction. You may see math.stackexchange.com/questions/1363393/… and let me know your comments. $\endgroup$– JeffCommented Jul 16, 2015 at 15:32
-
$\begingroup$ try $\gamma=3$ or $\gamma=4$, where the integral can be done exactly for finite $R$; then take the limit $R\rightarrow\infty$ and you get 1/2, what contradiction might there be? $\endgroup$ Commented Jul 16, 2015 at 20:15
-
$\begingroup$ solved, you are right. there is no contradiction. $\endgroup$– JeffCommented Jul 16, 2015 at 23:12