Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am attempting to find a closed form or symbolic expression of the inverse Fourier transform of the product of two Bessel functions of the first kind and a complex exponential, e.g.

$P(t) = IFT_w \{ F(w)\} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {J_0 (aw)J_0 (bw)e^{ - jcw} e^{jwt} dw}$

where $a, b, c$ are real values and $0<a, b, c<1$.

I've been tried so hard to solve it and recently found some clue in the textbook, Table of Integrals, Series, and Products, 7th ed. witten by Gradshteyn and Ryzhik:

$\int_0^\infty {e^{ - ax} J_v (\beta x)J_v (\gamma x)} dx = \frac{1}{{\pi \sqrt {\gamma \beta } }}Q_{v - \frac{1}{2}} \left( {\frac{{\alpha ^2 + \beta ^2 + \gamma ^2 }}{{2\beta \gamma }}} \right)$ , where $Q_{v}$ is Legendre function of the second kind,and the equality holds only if

${\mathop{\rm Re}\nolimits} \{ \alpha \pm i\beta \pm i\gamma \} > 0,{\rm }\gamma > 0,{\rm }{\mathop{\rm Re}\nolimits} \{ v\} > - \frac{1}{2}$.

Since the complex exponential term in $P(t)$ becomes $e^{-jw(c - t)}$; thus, $\alpha=j(c-t)$, and we also have $\beta=a$, $\gamma=b$, which are all reals, $P(t)$ does not satisfy ${\mathop{\rm Re}\nolimits} \{ \alpha \pm i\beta \pm i\gamma \} > 0$, so I couldn't use the result.

I've tried to get $P(t)$ with Mathematica, but it has returned with nothing. Today, I tried to solve it using the convolution theorem, but the result is somewhat very long and messy; the direct solving of the integral seems a nightmare for me.

Does anyone knows how to solve the integral or a closed form expression of $P(t)$? Any little clue will be helpful. Thanks in advance!

share|improve this question
    
Do you still need the answer? I think I computed something similar a while ago using convolution. If I am not wrong the result is compactly supported with logarithmic singularities somewhere. –  lcv Jun 11 '13 at 9:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.