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Let us define the following integral : $$W_n(r)=r\int_0^{+\infty} J_1(rt)[J_0(t)]^n dt,$$ with $r>0$ a real number and $n\in\mathbb{N}$ and where $J_0(x)$ and $J_1(x)$ are Bessel functions of the first kind. From Gradshteyn-Rizhik (2007 ed) formulas 6.511.1, 6.512.3, 6.513.9 we have : $W_0(r)=1$ for $r>0,$ $W_1(r)=1$ for $r>1$ and $W_2(r)=1$ for $r>2.$

My first question is : do we have $W_n(r)=1$ for $r>n$ ? I did some numerical verification using quadosc package in Python that tend to confirm this is true (I checked until $n=10$ with $r=n+1$).

My second question is : does it exist some closed formula for $W_n(r)$ when $r\le n$ ?

Thanks in advance for your help

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The answer to the first question is yes. According to Watson, "Treatise on the Theory of Bessel Functions" (1922), section 13.46, equation (8), $$ W_n'(r) = \int_0^\infty rt\, J_0(r t)\, [J_0(t)]^n \mathrm{d}t = 0\qquad \text{for }r > n.$$ Combined with $W_n(r) = \int_0^\infty J_0(x)\, [J_0(x/r)]^n \mathrm{d}x \to \int_0^\infty J_0(x)\, \mathrm{d}x = 1$ as $r\to\infty$, this shows that $W_n(r)=1$ for $r>n$.

Added: $W_n'(r)$ is the density of the norm of a sum of $n$ independent uniform unit-length vectors in the plane, so it is clear why $W_n'(r) = 0$ for $r>n$. This goes back to Kluyver, "A local probability problem", Koninklijke Nederlandsche Akademie van Wetenschappen, Proceedings, 8 (1906), 341–350.

For recent progress see the following paper (and references therein):

See Borwein, J.M., Straub, A., Wan, J., Zudilin, W. and Zagier, D., 2012. "Densities of short uniform random walks." Canadian Journal of Mathematics, 64(5), pp.961-990.

Added: In fact, $W_n(r) = P_n(0;r)$ in Borwein, Jonathan M. "A short walk can be beautiful." Journal of Humanistic Mathematics 6, no. 1 (2016): 86-109. The observation by Carlo Beenakker that $W_n(1) = \frac{1}{n+1}$ is attributed to Kluyver, see Example 2.3.

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  • $\begingroup$ Thanks a lot for your answer. Would you have any hint about the second question ? $\endgroup$
    – fbrx
    Commented Jun 18, 2021 at 11:07
  • $\begingroup$ A closed-form expression seems unlikely. See Section 13.48 of Watson's book (in particular the last display), where the large-$n$ asymptotics are discussed. $\endgroup$ Commented Jun 18, 2021 at 11:13
  • $\begingroup$ numerics indicates $W_n(n)=1$ for $n>1$ and $W_n(1)=1/(n+1)$ for $n\geq 1$ $\endgroup$ Commented Jun 18, 2021 at 12:34

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