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I am trying to solve a problem part of which includes the following integral ($j=\sqrt{-1}$):

$$\int_{k_1}^{k_2} k e^{-jk\sigma} J_n(\rho k) \, \mathrm{d}k$$

The $e^{-jk\sigma}$ term is making my usual approach to the problem ineffective. However, unless I am mistaken, this same term makes the above a forward Fourier transform where $\sigma$ would be the frequency term and $k$ the time/space term:

$$\rho^{-1} \int_{k_1}^{k_2} \rho k e^{-jk\sigma} J_n(\rho k) \, \mathrm{d}k = \rho^{-1} \mathcal{F}_\sigma \left( \rho k J_n(\rho k) \right) $$

I've been looking for this transform pair (i.e., $\mathcal{F}_x(xJ_n(x))$ but have not been able to find anything. Is there a analytic solution to this?

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1 Answer 1

I don't know if this helps but first try solving

$I(\sigma):=\int e^{-jk\sigma} J_n(\rho k)dk$

it is probably easier. Then use the fact that your integral above is simply $j\frac{dI}{d\sigma}$.

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It does and it doesn't in that I can get a solution for an integral with infinite limits but not for one with finite limits, unless I include convolution with a sinc to account for it. –  AnonSubmitter85 Nov 13 '12 at 4:05
    
you can extend the finite limits to infinite limits if you multiply the integrand by two opposite heaviside functions (the pulse function) –  Dar Far Nov 13 '12 at 12:04
    
Could you please rephrase this? I don't see how multiplying by the pulse function accomplishes this. Would the sifting property not be the result of multiplying the integrand by a pulse function? –  AnonSubmitter85 Nov 13 '12 at 22:36
    
well, the pulse function ${\bf 1}_{[k_1,k_2]}$ is defined over all $\mathbb{R}$ and has compact support in the interval you are interested in $[k_1,k_2]$. So, you can use this $\int_{-\infty}^{\infty}{\bf 1}_{[a,b]}f(x)dx=\int_{a}^{b}f(x)dx$ to extend your integral to the whole $\mathbb{R}$ and then use fourier transforms. Of course, when you take the derivative of the pulse function, it must be in a distributional sense (some dirac deltas will appear). –  Dar Far Nov 16 '12 at 12:18

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