1
$\begingroup$

I have the following equation: $I(k,x)=\int_0^xJ_k(\tau^k)d\tau=\alpha$ where $\alpha$ is a given constant $\alpha\in \mathbb{R}$ and $k$ integer with $k\gt 0$. $J_k(x)$ is the Bessel function of first kind. My question is: is it possible to solve the equation for every $k$ or there is only a couple $(k_0,x_0)$, solving the $I(k,x)=\alpha$?

Thanks in advance for suggestions

1 Answers 1

2

For $\alpha \in \mathbb{R}$ this can't be solved for every $k$. A simple counter-example shows this.

Consider $\alpha > 1 - J_{0}(j_{1,1})$ and $k = 1$.

Then: $ \int_{0}^{x} J_{1}(\tau) d\tau = 1 - J_{0}(x) \leq 1 - J_{0}(j_{1,1}) < \alpha.$ This integral is always smaller than $\alpha$, regardless of $x$.

  • 1
    Sorry, figured you would have seen the notation: $j_{n,k}$ is the $k$-th root of $J_{n}(x)$. Then $j_{1,1}$ is the first root of $J_{1}(x)$, and the first root of $J_{1}(x)$ coincides with the global minimum of $J_{0}(x)$ (http://www.wolframalpha.com/input/?i=plot+%5BBesselJ%5B0%2Cx%5D%2CBesselJ%5B1%2Cx%5D%5D).2012-03-09