Integral of product of spherical Bessel function of first kind with the second

Question

Notation: $j_l(x)$ and $y_l(x)$ denote spherical Bessel functions of the first and second kind, respectively.

I need a closed-form expression for the following indefinite integral: $$\int{x^2j_l(x)y_l(x)dx}$$ What I have tried: using the spherical Bessel differential equation: \begin{equation} x^2\frac{d^2j_l}{dx^2} + 2x\frac{dj_l}{dx} - \left(l\left(l+1\right)\right)j_l = -x^2j_l \end{equation} \begin{equation} x^2\frac{d^2y_l}{dx^2} + 2x\frac{dy_l}{dx} - \left(l\left(l+1\right)\right)y_l = -x^2y_l \end{equation} Multiplying the first by $y_l$, second by $j_l$, taking integral on both sides, adding and applying product rule on the left-most integral:

\begin{equation} x^2(j_ly_l)'|_{x_1}^{x_2} - 2\left(l\left(l+1\right)\right)\int j_ly_ldx= -2\int x^2j_ly_ldx \end{equation}

So I now need to know: $$\int j_ly_ldx $$ which I don't know how to proceed with and could find no references. Could someone please help me with the integral? Thank you.

Answer 1

We can use this gadget to compute the integral, although we have to set it up carefully: $j_l(sx)$ and $y_l(sx)$ satisfy the Sturm–Liouville equation $$ -\frac{d}{dx} \left( x^2 \frac{dy}{dx} \right) +l(l+1)y = s^2x^2 y, $$ so the gadget gives $$ \int x^2 j_l(sx)y_l(x) \, dx = -x^2\frac{j_l'(sx)y_l(x)-j_l(sx)y_l'(x)}{s^2-1} + C.$$

To find the integral when $s = 1$, we have to take the limit. The series expansion about $s=1$ can be computed using the Taylor expansions to be $$ \frac{x^2(j_l'(x)y_l(x)-j_l(x)y_l'(x))}{2(s-1)} + \frac{x^2}{4} (j_l'(x)y_l(x)-j_l(x)y_l'(x)) + \frac{x^3}{4}\big(j_l''(x)y_l(x)-j_l'(x)y_l(x)\big) + O(s-1) $$ However, the first terms are just constant because the Wronskian of $j_l(x)$ and $y_l(x)$, $W = j_l'(x)y_l(x)-j_l(x)y_l'(x)$, is proportional to $1/x^2$, so we can subtract off a function of $s$ so that the indefinite integral is continuous in $s$ at $s=1$. We hence obtain $$ \int x^2 j_l(sx)y_l(x) \, dx = \frac{x^3}{4}\big(j_l''(x)y_l(x)-j_l'(x)y_l(x)\big) + C', $$ which we can rewrite by using the recurrence relations and the differential equation if desired.

Answer 2

You probably can't find a nice trick to do this integral because the answer is quite ugly. For particular small integer values of $l$ we have: $$ I(l;x) \equiv \int x^2 j_l(x)y_l(x)dx \\ I(0;x) = -I(-1;x) = -\frac{\sin^2 x}{2} \\ I(1;x) = -I(-2;x) = -\frac{3x + x \cos(2x) - 2 \sin(2x)}{4x} \\ I(2;x) = -I(-3;x) = -\frac{5x^3 + x \cos(2x) (x^2-12) -6(x^2-1)\sin(2x)}{4x^3} \\ I(3;x) = -I(-4;x) = -\frac{7x^5 + x \cos(2x) (x^4-60x^2+180) -6(2x^4-25x^2+15)\sin(2x)}{4x^5}\\ I(4;x) = -I(-5;x) = -\frac{9x^7 + x \cos(2x) (x^6-180x^4+2940x^2-6300) -10(2x^6-93x^4+567x^2-315)\sin(2x)}{4x^7} $$ If you recognize those polynomials multiplying sine and cosine of $2x$ you may get a nicer formula, but I don't recognize those polynomials.