I am a bit confused about how to tackle the following integration and was hoping to find some help here.
Problem
I have an integral of the following form $$ \int_{x_i}^{x_f}\! f(a(x),y)\cdot j_l\left(y\left(x_f-x\right)\right)\,\mathrm{d}x $$ where $j_l$ is a spherical Bessel function.
Now, I'd like to re-express the whole integral depending on the variable $a(x)$ with $$ \frac{\mathrm{d}x}{\mathrm{d}a}=\frac{1}{a^2g(a)}, $$
assuming that the function $f(a(x))$ and $g(a)$ are known as well as the limits of the integration $a(x_f)=1$ and $a(x_i)=0$.
Question
How can I get rid of the $x$-dependence and re-express this as an integral over $a$? So, mainly, what do I have to do with the spherical Bessel function?
Attempt
I see that the argument in the Bessel function, $x_f-x$ can be expressed as $\int_{a(x)}^{a(x_f)}\!\frac{\mathrm{d}a}{a^2g(a)}$, but I think the result is not just
$$ \int_{a(x_i)}^{a(x_f)}\! f(a,y)\cdot j_l\left(y\left(\int_{a(x)}^{a(x_f)}\!\frac{\mathrm{d}a}{a^2g(a)}\right)\right)\,\frac{\mathrm{d}a}{a^2g(a)}, $$ but that there should be an additional factor or dependence arising from the change of variable in the Bessel function, but I cannot wrap my head around what it should be.
If you'd spot what it is, you'd be my personal hero of the day. Thank you very much in advance!
$\dfrac{\mathrm d x}{\mathrm d a}(a) =\dfrac 1{a^2\,g(a)}$ entails that $x(a) = \displaystyle\int \dfrac{\mathrm d a}{a^2~g(a)}~+C$, where $x$ is a function of $a$, and $C$ is an arbitrary constant.
So the term you want is this:
$$x_f-x(a) = x_f-\int\dfrac{\mathrm d a}{a^2~g(a)}-C$$