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Let$(X,\mu)$ be a measure space. $f:X \to \mathbb{R}$ be a measure function. For every $t\in \mathbb{R}$ the distribution function $F$ of $f$ is defined as $ F(t)=\mu\{x \in X:f(x)<t\}.$

I have difficulties of finding distribution function of the bessel function of the first order, i.e. $F(t)=\{x \in \mathbb{R}: f(x)=\frac{|J_1(x)|}{x}<t\}$. Any ideas or references will be very helpful. Thank you.

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The question is not clear. Do you mean a cumulative distribution function? – wnvl Jan 24 '12 at 21:53
The cumulative distribution function you are defining is generally used in probability theory, where the space has finite total measure (and is used to convert a random variable on a complicated space to something on $\mathbb R$). If you aren't working on a finite measure space, you need some argument that $F(t)<\infty$ for any value of $t$. The graphs on wikipedia make $J_1(x)$ look similar to a dampened harmonic oscilator, and if that is the case, a local extrema at $x_0$ is a global extrema on $[x_0,\infty)$, and so $F(J_1(x_0))=\infty$. – Aaron Jan 25 '12 at 0:21
Sorry, I've lost denominator. Now we can consider $F$ on $(0,1)$ – Nick G.H. Jan 25 '12 at 0:42
up vote 0 down vote accepted

$J_1(x)/x$ is an even function whose global maximum value is $1/2$ at $x=0$ (actually that's a removable singularity). The global minimum value is approximately $-0.06613974369805002$, attained at approximately $x = \pm 5.135622301840683$. It's unlikely that there are "closed-form" expressions for these numbers, or for any $x$ for which $J_1(x)/x$ is rational. So I don't know what you expect an answer to look like.

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Hm. We can consider even $2\frac{|J_1|}{x}$ to make interval $(0,1)$. But now, can We at least say simething about derivative of F? If I understand right, it is $F'(t)=\sum\frac{1}{f'(x)}$? – Nick G.H. Jan 25 '12 at 2:18
Also, if to consider measure something like $d\mu(x)=1/x^r$, can be F found? – Nick G.H. Jan 25 '12 at 2:29
If $\mu$ is a finite measure with a continuous density, so $d\mu =g(x)\ dx$ where $g$ is continuous, and $F(t) = \mu(\{x: f(x) < t\})$ where $f$ is continuously differentiable, then at any $t$ for which $f(x)=t$ has finitely many solutions $x_j$, at all of which $f' \ne 0$, we should have $F'(t) = \sum_j g(x_j)/|f'(x_j)|$. – Robert Israel Jan 25 '12 at 7:43

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