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

$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^{-\gamma}}} \mathrm{d}r$$f(\omega) = \lim_\limits{R \to \infty} \frac{1}{R^2} \int_{r=0}^{R}{re^{ \omega r^{-\gamma}}} \mathrm{d}r$, where all parameters are constant except $i$ which is the imaginary unit, and $\gamma > 2$.

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

$f(\omega) = \lim_\limits{R \to \infty} \frac{a}{R^2}\left( 1 - (1+bR)e^{-cR}\right) \int_{r=0}^{R}{re^{i \omega r^{-\gamma}}} \mathrm{d}r$, where all parameters are constant except $i$ which is the imaginary unit, and $\gamma > 2$.

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 $\gamma > 2$.

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Jeff
Jeff
  • 482
  • 2
  • 8

Simplifying an expression using tools from Fourier transform

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

$f(\omega) = \lim_\limits{R \to \infty} \frac{a}{R^2}\left( 1 - (1+bR)e^{-cR}\right) \int_{r=0}^{R}{re^{i \omega r^{-\gamma}}} \mathrm{d}r$, where all parameters are constant except $i$ which is the imaginary unit, and $\gamma > 2$.