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I would like to solve the following integral

$$ \int_0^\infty e^{-a k^2} J_{3/2}(b k) J_{3/2}(c k) J_{3/2}(f k) J_{1/2}(r k) k^{-3} dk, $$

where $a,b,c,f,r > 0$, and $J_\nu(x)$ is the Bessel function of order $\nu$.

An equivalent (within proportionality) integral in terms of spherical Bessel functions is

$$ \int_0^\infty e^{-a k^2} j_1(b k) j_1(c k) j_1(f k) j_{0}(r k) k^{-1} dk, $$

So far I haven't found the integral in any integration tables. Any guidance on how to solve it would be most appreciated!

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1 Answer 1

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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 functions) are discussed in Fabrikant - Elementary exact evaluation of infinite integrals of the product of several spherical Bessel functions, power and exponential, where the main idea is to use the following identity: \begin{align} I(p, q, i, j, k, l; a, b, c, d) &= (-1)^{i+j+k+l} a^i b^j c^k d^l \frac{\partial^i}{(a \partial a)^i} \frac{\partial^j}{(b \partial b)^j} \frac{\partial^k}{(c \partial c)^k} \frac{\partial^l}{(d \partial d)^l} \biggl[\\ &\int_0^\infty dt \exp(-p t^2) \frac{ j_0(a t) j_0(b t) j_0(c t) j_0(d t) } { t^{i + j + k + l - q} } \biggr]. \end{align}

The key point is to now expand the zeroth order spherical Bessel functions into trigonometric functions, and converting the products of the trigonometric functions into sums: \begin{align} \sin(ax) \sin(bx) \sin(cx) \sin(dx) =& \frac{1}{8} \biggl\{ \cos[(a + b + c + d)x] + \cos[(a + b - c - d)x] + \cos[(a - b + c - d)x]\\ &+ \cos[(a - b - c + d)x] - \cos[(-a + b + c + d)x] - \cos[(a - b + c + d)x]\\ &- \cos[(a + b - c + d)x] - \cos[(a + b + c - d)x] \biggr\} \end{align} followed by the use of the following integral, which is not considered in the reference above, but can be found in Gradshteyn and Ryzhik, 7th ed., formula 3.953.8: $$ \mathcal{I}(p, s; n) := \int_0^\infty dt\, t^n \exp(-p t^2) \cos(s t) = \frac{1}{2} p^{\frac{-(n + 1)}{2}} \, e^{-s^2 / 4 p} \Gamma \left(\frac{1}{2} + \frac{n}{2}\right) \, _1F_1\left(-\frac{n}{2}; \frac{1}{2}; \frac{s^2}{4 p}\right). $$

Note that the formal requirement is that $\operatorname{Re}(n) > -1$, but the above result can be understood as an analytic continuation for general values $p, s, n$.

Additionally, it can happen that one of the "angles" above is zero, in which case we have the integral: $$ \mathcal{I}(p, 0; n) := \int_0^\infty dt\, t^n \exp(-p t^2) = \frac{1}{2} p^{-\frac{n}{2}-\frac{1}{2}} \Gamma \left(\frac{n+1}{2}\right) $$ with the same condition on $n$ as above.

The result in your specific case is then: \begin{align} I(p, -1, 1, 1, 1, 0; a, b, c, d) &= - a b c \frac{\partial}{(a \partial a)} \frac{\partial}{(b \partial b)} \frac{\partial}{(c \partial c)} \int_0^\infty dt \exp(-p t^2) \frac{ j_0(a t) j_0(b t) j_0(c t) j_0(d t) } { t^4 }\\ &= - a b c \frac{\partial}{(a \partial a)} \frac{\partial}{(b \partial b)} \frac{\partial}{(c \partial c)} \int_0^\infty dt \exp(-p t^2) \frac{ \sin(a t) \sin(b t) \sin(c t) \sin(d t) } { a\, b\, c\, d\, t^8 }\\ &= - \frac{\partial}{\partial a} \frac{\partial}{\partial b} \frac{\partial}{\partial c} \bigg[ \frac{1}{a\, b\, c\, d} \int_0^\infty dt \exp(-p t^2) \frac{ 1 } { t^8 } \frac{1}{8} \bigg\{ \\ &\cos[(a + b + c + d)t] + \cos[(a + b - c - d)t] + \cos[(a - b + c - d)t] \\ &+ \cos[(a - b - c + d)t] - \cos[(-a + b + c + d)t] - \cos[(a - b + c + d)t] \\ &- \cos[(a + b - c + d)t] - \cos[(a + b + c - d)t] \bigg\} \bigg] \\ &= - \frac{1}{8} \frac{\partial}{\partial a} \frac{\partial}{\partial b} \frac{\partial}{\partial c} \biggl\{\frac{1}{a\, b\, c\, d} \\ &\mathcal{I}(p, a + b + c + d; -8) + \mathcal{I}(p, a + b - c - d; -8) + \mathcal{I}(p, a - b + c - d; -8) \\ &+ \mathcal{I}(p, a - b - c + d; -8) - \mathcal{I}(p, -a + b + c + d; -8) - \mathcal{I}(p, a - b + c + d; -8) \\ &- \mathcal{I}(p, a + b - c + d; -8) - \mathcal{I}(p, a + b + c - d; -8) \biggr\}. \end{align}

The explicit result is fairly cumbersome to fully write out; below is an example Mathematica code which can be used as a starting point to generate the full solution (when $a \pm b \pm c \pm d \neq 0$) and compare it with the numerical result:

numeric[p_, a_, b_, c_, d_] := NIntegrate[
  Exp[-p t^2] SphericalBesselJ[1, a t] SphericalBesselJ[1, 
    b t] SphericalBesselJ[1, c t] SphericalBesselJ[0, d t]/t,
  {t, 0, Infinity}
  ];
integral[p_, s_, n_] := 
  1/2 p^(-(n + 1)/2) Exp[-s^2/(4 p)] Gamma[
    1/2 + n/2] Hypergeometric1F1[-n/2, 1/2, s^2/(4 p)];
result = -1/8 Table[
    Series[
       expression,
       {epsilon, 0, 0}
       ] // Normal // D[#/(a b c d), a, b, c] &,
    {
     expression,
     {
      integral[p, a + b + c + d, -8 + epsilon],
      integral[p, a + b - c - d, -8 + epsilon],
      integral[p, a - b + c - d, -8 + epsilon],
      integral[p, a - b - c + d, -8 + epsilon],
      -integral[p, -a + b + c + d, -8 + epsilon],
      -integral[p, a - b + c + d, -8 + epsilon],
      -integral[p, a + b - c + d, -8 + epsilon],
      -integral[p, a + b + c - d, -8 + epsilon]
      }
     }
    ] // Total;

No idea if the solution which the code above generates can be simplified though.

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  • $\begingroup$ I think the GR reference should be formula 3.952.8. Great find with Fabrikant! $\endgroup$
    – ste
    Mar 11 at 17:00

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