3
$\begingroup$

I know there is a $\delta$-function representation in terms of the Bessel function as

$$\delta(x-a) = x\int_0^{\infty} t J_{\nu}(xt)J_{\nu}(at)\; dt$$

Is there something similar for

$$\int_0^{\infty} t J_{\mu}(xt)J_{\nu}(at)\; dt$$

Maybe there is something I can do to at least get a reasonable series representation... All I can find are the integrals where the power of $t$ is related in a certain way to the orders $\mu$ and $\nu$, but it is not the case in my case. Thanks!

$\endgroup$
0

2 Answers 2

1
$\begingroup$

In general, $\mu$ and $\nu$ are not equal, representation in the form of series can be obtained if you replace the product of only Bessel function under the integral by expansion in series where each term is expressed via hypergeometric function $_2F_1$ (see Bateman and Erdelyi, Higher Transcedental Functions, Volume $2$, Chapter $7 (7.2.7$). Then you can integrate term by term.

$\endgroup$
0
$\begingroup$

The integral you seek is null for $\mu\ne\nu$. See the book by Smythe, "Static and dynamic electricity", sections 5.296-5.298.

$\endgroup$
1
  • $\begingroup$ I'm pretty sure that this statement is wrong. For $\mu \neq \nu$ the integrand is "just" some rapidly oscillating function. You can evaluate it numerically, for example by integrating against $e^{-\epsilon t}$ for fixed \epsilon and letting $\epsilon \to 0$, to see that it has a finite limit. The null result from the book you're quoting involves integrals of the form $\int_0^1\!dx\, x\, J_n(k_1 x) J_n(k_2 x) = 0$ where $k_{1,2}$ are the eigenvalues of Bessel's equation that align with a specific boundary condition at $x=1$. Crucially, the two Bessel's have identical $n$ i.e. $\mu$. $\endgroup$ Mar 28, 2023 at 10:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .