Suppose I have that: $$ \int_0^\pi xJ_0(\alpha x)g(x) dx \geq 0 \ \ \ \ (*) $$ then is there anything I can say about: $$ \int_0^\pi xJ_{\frac{-1}{2}}(\alpha x)g(x) dx \ \ (\geq 0?) \ \ \ (**)$$

I understand that this will heavily depend on $g(x)$, in fact, because of $(*)$ we see that $g(x)$ is a positive definite radial function on $R^2$.

Suppose we impose the condition of non-increasing, and non-negative on $g(x)$, would $(*) \geq 0 $ imply $(**) \geq 0 $.

I thought of this because $J_0$ and $J_{\frac{-1}{2}}$ have very similar shape except at $x=0$. Similarly, $J_1$ and $J_{\frac{1}{2}}$ also have very similar shape (just different scaling)... J0 and J-1/2

For the picture above, the blue curve is $J_{\frac{-1}{2}}$, and the orange curve is $J_{0}$.

Also the plot of $J_{\frac{1}{2}}$ (Blue curve) and $J_1$ (Orange curve) is as follow: enter image description here

If you can direct me to sources where I can read and study more about these things, then that would be great too.

  • $\begingroup$ Since $J_{\mp\frac{1}{2}}(\alpha x)$ are just decaying $\cos(\alpha x)$ and $\sin(\alpha x)$ respectively, you can probably use quite a bit of Fourier Theory to make statements about (**). $\endgroup$ – Andy Walls May 9 at 12:22

Long Comment:

One quick and easy starting point is to connect them by using the approximation

$$J_n(x)\approx\sqrt{\frac{2}{\pi \left|x\right|}} \cos \left(-\frac{ n\pi }{2}+\left|x\right|-\frac{\pi }{4}\right)$$

Actually for $n=\pm\, \frac{1}{2}$ this is exact which helps a lot. (See Bessel Functions - Asymptotic forms)

The graph comparing $J_0(x)$ (Blue) with it's approximation (Orange) is

enter image description here

Accuracy is only seriously lost close to the y-axis, for $(0\le |x|<\pi))$. You could make a spline fit using a different function in this range to improve the fit for "engineering" rather than pure mathematics purposes; otherwise some deeper analysis using Fourier theory will be needed.

In terms of your integral the amount of effort you need to put in to improve the fit depends on the value of $\alpha$ you use - and whether or not (in the greater scheme of things) it is a constant or a variable.

In terms of answering your question: Suppose we impose the condition of non-increasing, and non-negative on $g(x)$, would $(∗)≥0$ imply $(∗∗)≥0$?

I'm not sure. It may help as a first step to use the substitution $y=\alpha x$ to rewrite your integrals, so you can separate the problem into the two ranges $0\le |y| \le \pi$ and $\pi < |y| \le \alpha\pi$.

You can then modify your question to: Is the positive result from the first range always greater than any negative result from the second range?


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