25

See Densities of Short Uniform Random Walks (with an appendix by Don Zagier) by
Jonathan M. Borwein, Armin Straub, James Wan, and Wadim Zudilin,
Canad. J. Math. Vol. 64 (5), 2012 pp. 961–990. http://cms.math.ca/10.4153/CJM-2011-079-2
Your integral is $p_4(1)$ in Equation (2.1). They write it in terms of hypergeometric functions (doing this for $p_4$ is one ...

16

In fact, precisely your integral has been computed in closed form, in: Annie Gervois and Henri Navelet, Some integrals involving three Bessel functions when their arguments satisfy the triangle inequalities, J. Math. Phys. 25 (1984), no. 11, 3350–3356. Their result is
$$
\int_0^\infty J_m(ar)J_n(br)J_{m+n}(cr)r\,dr =
\begin{cases}
0&\text{if }c^2 < (a-...

answered Jan 15 '14 at 22:43

Francois Ziegler

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13

The Bessel functions $J_\ell$ for $\ell\geq 1$ odd are pairwise orthogonal on the positive axis with respect to the measure $dx/x$. They correspond to the holomorphic spectrum (of various even weights $\ell+1$) of $L^2(\Gamma\backslash H)$. The orthogonal complement of the span of these $J_\ell$'s is continuously (and orthogonally) spanned by the functions $...

9

This triple-Bessel integral was studied in much detail by S.K.H. Auluck, On the Integral of the Product of Three Bessel Functions over an Infinite Domain. Closed form expressions are given if two the three orders coincide. For the more general case, a reliable way to evaluate this integral using Mathematica is described and tested. The numerical difficulties ...

answered Jan 13 '14 at 5:37

Carlo Beenakker

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9

The paper here mentions the integral $\displaystyle\sum_{n = -\infty}^\infty J_n^4(x) = \frac{2}{\pi} \int_0^\frac{\pi}{2} J_0^2(2x \sin \theta) d\theta$
and the asymptotic $\displaystyle \sum_{n = -\infty}^\infty J_n^4(x) = \frac{1}{x\pi^2}(\log x + 5 \log 2 + \gamma) + O(x^{-\frac{3}{2}})$ as $x \to \infty$, and gives references.

7

juan's suggestion of the Hankel function seems great.
Here is $J_0(x)$ and $u(x):=|J_0(x)+iY_0(x)|$.
Of course $u(x) \ge J_0(x)$ everywhere and $u(x) = |J_0(x)|$ at all the zeros of $Y_0(x)$. For asymptotic properties of $u(x)$, from Maple I get asymptotics
$$
u(x) = \sqrt{\frac{2}{\pi x}}+O(x^{-3/2})\qquad\text{as } x \to \infty
$$
which explains the good ...

7

I was acquainted with Nikolay Vasil'evich Kuznetsov while worked in Vladivostok, 1990s. And he was very kind to mee, too. He tought me that many asymptotics for Bessel functions are not valid, many from Erdelyi's book on asymptotics, and some formulas for integral transforms. And as you I do not remember all his lessons, unfortunately.
But I think the ...

6

This integral is in the literature (e.g. Bateman manuscript project Vol. 2, Equation 8.9(5)).
$ \int_0^\infty x^\mu \: L_n^{(\mu)}(\alpha x^2) \: J_\mu(xy) \: \mathrm{e}^{-\beta x^2} \: \mathrm{d}x = 2^{-\mu-1} \: \beta^{-\mu-n-1} \: (\beta - \alpha)^n \: y^\mu \: \mathrm{e}^{-y^2/(4\beta)} \: L_n^{(\mu)}\left(\frac{\alpha y^2}{4\beta(\alpha - \beta)}\right)...

6

The integral can be expressed in terms of Bessel and Struve functions,
$$I(\xi)=\int_0^{\infty} dq\, \frac{J_0(q \xi)}{q+1}=\tfrac{1}{2} \pi \bigl(\pmb{H}_0(\xi)-Y_0(\xi)\bigr).$$
The small-$\xi$ behavior is
$$I(\xi)=\xi-\ln \xi-\gamma_{\rm Euler} +\ln 2 +{\cal O}(\xi^2).$$
For large $\xi$ one has
$$I(\xi)=1/\xi-\tfrac{1}{16}\sqrt\pi\xi^{-3/2}(\cos\xi+\sin\...

5

I don't see a relation between the two. But your first identity can be deduced from the relation
$$
\sum_{n=-\infty}^\infty \frac{J_n(y)^2}{n+x}=\frac\pi{\sin(\pi x)}J_x(y)J_{-x}(y)
\tag{2.3}
$$
established by M. D. Rogers (2005) (his equation numbering). In fact, putting $x=\frac12$ and using $J_{1/2}(y)=\left(\frac2{\pi y}\right)^{1/2}\sin y$ and $J_{-1/2}(...

answered Sep 13 '15 at 9:36

Francois Ziegler

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5

Watson? (1922, pp. 486, 516, 521) has for the smallest positive zero:
$$
\sqrt{\nu(\nu+2)}<x_\nu<\sqrt{2(\nu+1)(\nu+3)},
$$
$$
x_\nu= \nu+1.855757\nu^{1/3}+O(\nu^{-1/3}).
$$

answered Nov 17 '18 at 19:56

Francois Ziegler

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5

From the definition 10.47.10, it follows that
$$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = h_n^{(1)}(z)\cdot h_n^{(2)}(z).$$
So, by the expansions 10.49.6 and 10.49.7,
\begin{split}
\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) &= \sum_{k=0}^n I^{k-n-1} \frac{a_k(n+\frac{1}{2})}{z^{k+1}}\cdot \sum_{l=0}^n (-I)^{l-n-1} \frac{a_l(n+\frac{1}{2})}{z^{l+1}} \...

5

The coefficient of $x^n$ in $_2F_0(\alpha,\beta;z) {}_2F_0(\alpha,\beta;-z)$, multiplied by $n!$,
is
$$\begin{aligned}
\sum_{k=0}^n (-1)^{n-k}\binom nk &(\alpha)_k(\beta)_k (\alpha)_{n-k}(\beta)_{n-k}\hfill\\
&=(-1)^n (\alpha)_n(\beta)_n\,
{}_3F_2\left({-n,\atop}\!{\alpha,\atop 1-\alpha -n,}\,
{\beta\atop1-\beta-n}\biggm | 1\right).
\end{aligned}...

4

Maybe you can accept the following result as a closed form. Transforming the variable of integration in your integral
$$
I_{n}(a,b):= \frac{1}{2 a}\ e^{-a/2}\int_{0}^{\infty} dx\ x^{(b-2)/4} e^{-x/2}(1-e^{-x/2})^{n} I_{\frac{b}{2}-1}(\sqrt{a x})
$$
to $y:=\sqrt{a x}$, we get
$$
I_{n}(a,b)=e^{-a/2}\ a^{-\frac{b}{4}-\frac{3}{2}} \int_{0}^{\infty}dy\ y^{b/2}\ ...

4

Although I'm sure @Zurab's answer will indeed give me a solution (and I've heard the suggestion before in other situations), I'm not very familiar with the technique.
Following the suggestion by @Johannes in the comments, there is a much more straightforward way that, although not fully general, fits my problem exactly (i.e. it leads to expressions for the ...

4

According to Uniform Upper and Lower Bounds on the Zeros of Bessel Functions of the First Kind, p 2:
$$ j_{v,k} > v + \frac23 |a_{k-1}|^{3/2}$$
where $j_{v,k}$ is the $k$-th positive zero of $J_v$.
So the answer is no, since the first zero is greater than one.

4

Following on from the answer of Sam Dolan, I generalized my conjecture to the following, which I prove as Theorem 4 in my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials. [The statement in the question is case where $j=0$.]
Theorem. For $0\le j \le p$, $\mathbf{s}\in \mathbb{R}^{2p+1}$, with $s=|\mathbf{s}|$ and $R&...

4

I have proved this Bessel identity in my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials. It is Theorem 4 which is proved in Section 3.

4

The Bessel functions have the series expansion
$$ J_n(z) = \sum_{k=0}^\infty \dfrac{(-1)^k (z/2)^{n+2k}}{k! (k+n)!} $$
If $0 < z \le 2 \sqrt{1+n}$ this is a convergent alternating series with positive initial term and terms nonincreasing in absolute value (strictly decreasing after the first), implying that the sum is greater than $0$.

4

The relationship has actually motivated several studies:
Eisenstein series and the Riemann zeta function (1981)
Moments of the Riemann zeta function and Eisenstein series part I and part II (2004)
The Riemann hypothesis for
certain integrals of Eisenstein series (2004)
Some of its limitations are discussed in an answer to this MO question.

answered Jun 13 '15 at 14:09

Carlo Beenakker

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4

By changing $s=e^{t-x}$ in the integral, one obtains $$I(t)=\int_1^\infty e^{-e^{-t}s-e^t/s}\frac{ds}{s}$$ which corresponds to the definition of the incomplete Bessel function in Harris:
$$ I(t)=K_0(e^{-t},e^t)$$ With eq. (38) in the article,
$$I(t)=K_0(2)+tI_0(2)+2\sum_{j=1}^\infty \frac{(-1)^j}{j}I_j(2)\sinh(jt)$$
where $I_n$ and $K_n$ are the modified ...

4

as requested by the OP in the comment section:
$$\int_0^T e^{-x}I_n(x)\frac{1}{x}\,dx=\frac{1}{n}+\frac{1}{n T^{n-1}}e^{-T}\left[a_n(T)I_0(T)+b_n(T)I_1(T)\right]$$
the functions $a_n$ and $b_n$ are polynomials of degree $n-1$, I do not have a closed form expression; the first few are:
$$a_1(T)=-1,\;\;a_2(T)=-2T,\;\;a_3(T)=-3 T^2+4 T,$$
$$a_4(T)=-4 T^3+8 ...

answered Jun 11 '17 at 16:44

Carlo Beenakker

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4

Mathematica can evaluate the integral$^\ast$
$$I_{n,l}=\begin{align}
\int_{0}^{\infty } e^{-\frac{r^2}{2B}} r^{l-n}
L_n^{l-n}\left(\frac{r^2}{C}\right) I_{l-n}\left(\rho r \right) r dr
\end{align},\;\;B,C,\rho>0,$$
for any integer $n\geq 0$ as a function of $l>n-1$. The results are consistent with
$$I_{n,l}=\begin{align}
B^{l+1} (1/B-2/C)...

answered Aug 12 '17 at 14:45

Carlo Beenakker

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4

The modified Bessel functions satisfy the recurrence
$$ I_n(x) - \frac{2(1+n)}{x} I_{n+1}(x) - I_{n+2}(x) = 0 $$
which translates to a first-order differential equation for
$g(\lambda) = \sum_{n=0}^\infty I_n(x) \lambda^n$ (note that I'm including $n=0$ in this sum):
$$ 2 \lambda^2 g'(\lambda) + (1-\lambda^2) x g(\lambda) = x I_0(x) + \lambda x I_1(x) $$
...

4

The integral is studied in On Certain Indefinite Integrals Involving Bessel Functions (1958).
(The $i=0$ integral is $g(a,0,x)$ in the notation of that paper, and as Robert Israel points out, the $i=1$ integral is simply related.)
The paper is behind a paywall, so I have not read it.
And then there is Tables of some indefinite integral of Bessel functions ...

answered Feb 12 '18 at 0:24

Carlo Beenakker

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4

The term $\frac{\pi}{2s}$ comes from the pole of $\zeta(s)$. Let's use the fact that Mellin transform of $K_0(s)$ equals
$$
\int_0^{+\infty} K_0(s)s^{t-1}ds=2^{t-2}\Gamma^2(t/2).
$$
From this we get that if $f(s)=\sum_{n} K_0(sn)$ then for $\mathrm{Re}\,s>1$ we have
$$
\int_0^{+\infty} f(s)s^{t-1}ds=2^{t-2}\Gamma^2(t/2)\zeta(t)
$$
Therefore, by Mellin ...

ca.classical-analysis-and-odes analytic-number-theory asymptotics computational-number-theory bessel-functions

4

In "Higher Trancendental Functions", Vol. 1, by A. Erdelyi (ed.), on page 86, equation (4) says
$$
_2F_0(α,β;z)\ _2F_0(α,β;-z) = \, _4F_1(α,β,\frac{1}{2}(α+β),\frac{1}{2}(α+β+1);α+β;4z^2),
$$
from which your formula can be derived by setting $\alpha=−n$, $\beta=n+1$.
Unfortunately Erdelyi, et al., do not give a proof only references, which I reproduce ...

4

Let $n=\mu - \nu$ be an integer, as specified in the problem. Then I'll indicate how to prove that
$$I:=\int_0^\infty e^{-a\,t} t^{\nu+1} J_\nu(b\,t) L_n^{2v}(t) \,dt =\frac{(2b)^\nu \Gamma(\nu+1/2)}{\sqrt{\pi}}\sum_{k=0}^n \frac{(-1)^k}{k!}\binom{n+2\nu}{n-k} $$
$$
(a^2+b^2)^{-(\nu+k+3/2)}a^{k+1}{}_2F_1(-k/2, -k/2-1/2,1+\nu,-(b/a)^2)(2\nu+1)_{k+1}$$
...

4

Maple gave me this...
$$
J_0(x) = \left( {\frac {\sin \left( x \right) }{\sqrt {\pi}}}+{\frac {\cos
\left( x \right) }{\sqrt {\pi}}} \right) x^{-1/2}+ \left( -{\frac {\cos \left( x \right) }{8\sqrt {\pi}}}+{\frac {\sin
\left( x \right) }{8\sqrt {\pi}}} \right) x^{-3/2}
\\+ \left( -{\frac {9\,\sin \left( x \right) }{128\,\sqrt {\pi}}}-{
\frac {9\,\cos \...

4

Let's consider the second integral, which can be written in the following form:
$$
I(p, q, i, j, k, l; a, b, c, d)
:=
\int_0^\infty
dt\,
\exp(-p t^2)
t^q
j_i(a t) j_j(b t) j_k(c t) j_l(d t)
$$
where in your case, $i = j = k = 1$, $l = 0$, and $q = -1$.
These kinds of integrals (as well their generalization to a product of arbitrarily many spherical Bessel ...

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