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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

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1 Answer 1

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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 $x$.

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what is $j_{1,1}$? –  Riccardo.Alestra Mar 8 '12 at 12:25
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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 $J_{0}(x)$ (wolframalpha.com/input/…). –  in_wolfram_we_trust Mar 9 '12 at 8:15

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