# Integral with 4 Bessel functions and an exponential

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!

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.

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