analytical approximation of an integral of a bessel function around a small range I an trying to find an analytical way to approximate (via power series expansion or otherwise)
$$\int_{d-\delta}^{d+\delta}  j_0\left( \alpha \sqrt{1-\frac{z}{\sqrt{\beta+z^2}}} \right) dz$$
The integration range is positive $d>0$,  and $d \gg \delta$ (not sure if factor 10 to 100 justify the $\gg$ sign though) . All parameters $\alpha$,$\beta$ etc  are positive,   the $j_0$ is the Spherical Bessel function, that has an analytical power series
$$j_0(x) = \frac{1}{2} \sqrt{\pi} \sum_{k=0}^{\infty}  \frac{( \frac{-1}{4})^k x^{2 k}}{k! \Gamma(3/2 + k)}
$$
but taking the first few terms is highly inaccurate and the gamma function is an issue to integrate.
Any insight on how to use the close integration limits to obtain a few term expansion?
 A: As @K.defaoite already answered, the idea is to make an expansion around $z=d$.
You problem being
$$y(z)=\frac{\sin (f(z))}{f(z)}$$ write
$$f(z)=f(d)+ f'(d)(z-d)+\frac{1}{2}  f''(d)(z-d)^2+\frac{1}{6} f^{(3)}(d)
   (z-d)^3+O\left((z-d)^4\right)$$ replace and continue with Taylor series to make
$$y=\frac{\sin (f(d))}{f(d)}+\frac{f'(d) (f(d) \cos (f(d))-\sin (f(d)))}{f(d)^2}(z-d)+$$
$$\frac{f(d) f''(d) (f(d) \cos (f(d))-\sin (f(d)))-f'(d)^2
   \left(\left(f(d)^2-2\right) \sin (f(d))+2 f(d) \cos (f(d))\right)}{2 f(d)^3}(z-d)^2+O((z-d)^3$$ and
$$\int_{d-\delta}^{d+\delta} y(z)\,dz=\frac{2   \sin (f(d))}{f(d)}\delta+
\Bigg[S \sin (f(d))+C \cos (f(d))  \Bigg]\delta^3+\cdots$$ where
$$S=-\frac{f''(d)}{3 f(d)^2}-\frac{f'(d)^2}{3 f(d)}+\frac{2 f'(d)^2}{3 f(d)^3}$$
$$C=\frac{f''(d)}{3 f(d)}-\frac{2 f'(d)^2}{3 f(d)^2}$$
Computing the derivatives
$$S=\frac{\beta  \left(-6 d^3-\left(\alpha ^2-3\right) \beta  \sqrt{\beta +d^2}+6 d^2
   \sqrt{\beta +d^2}+\left(\alpha ^2-6\right) \beta  d\right)}{12 \alpha 
   \left(\beta +d^2\right) \left(\beta -d \sqrt{\beta +d^2}+d^2\right)^{5/2}}$$
$$C=-\frac{\beta }{4 \left(\beta +d^2\right)^2}$$
Just for a quick test with $\alpha=2$, $\beta=3$ , $d=5$ and $\delta=\frac 12$, the first term is
$$\frac{\sin \left(2 \sqrt{1-\frac{5}{2 \sqrt{7}}}\right)}{2 \sqrt{1-\frac{5}{2
   \sqrt{7}}}}= 0.9636766423$$ and the second is
$$\frac{-\sqrt{42 \left(2198+815 \sqrt{7}\right)} \sin
   \left(\sqrt{4-\frac{10}{\sqrt{7}}}\right)-126 \cos
   \left(\sqrt{4-\frac{10}{\sqrt{7}}}\right)}{1053696}=-0.0002902393$$ which is almost negligible (just as @K.defaoite already told).
All ot this makes a total of $\color{red}{0.96338}64$
to be compared with the numerical integration which leads to $\color{red}{0.9633844}$
A: Let's first do a theoretical analysis. Consider an integral of the form
$$F(s,\epsilon)=\int_{s-\epsilon}^{s+\epsilon}f(x)\mathrm dx$$
Using Taylor's theorem, in the vicinity of $s$,
$$f(x)=\sum_{k=0}^{2n}\frac{f^{(k)}(s)}{k!}(x-s)^k~+\mathrm O\big((x-s)^{2n+1}\big)$$
And so,
$$F(s,\epsilon)=\sum_{k=0}^{2n}\int_{s-\epsilon}^{s+\epsilon} \frac{f^{(k)}(s)}{k!}x^k\mathrm dx+\int_{s-\epsilon}^{s+\epsilon} \mathrm O\big((x-s)^{2n+1}\big)~\mathrm dx \\ =\sum_{k=0}^n \int_{s-\epsilon}^{s+\epsilon}\frac{f^{(2k)}(s)}{(2k)!}x^{2k}\mathrm dx+\mathrm O(\epsilon^{2n+2}) \\ =2\sum_{k=0}^n \frac{f^{(2k)}(s)}{(2k+1)!}\epsilon^{2k+1}+\mathrm O(\epsilon^{2n+2})$$
Choosing $n=0$ is essentially the very obvious approximation
$$\int_{-a}^a f(x)\mathrm dx\approx 2a~f(0)$$
And we can add as many orders as we like after that, but the $n=0$ order may already be enough for you.
Recalling that $j_0(x)=\sin (x)/x$, the zeroth order estimate is
$$\int_{s-\epsilon}^{s+\epsilon}j_0\left(\alpha\sqrt{1-\frac{x}{\sqrt{\beta+x^2}}}\right)\mathrm dx=2\epsilon~\frac{\sin\left(\alpha\sqrt{1-\frac{s}{\sqrt{\beta+s^2}}}\right)}{\sqrt{1-\frac{s}{\sqrt{\beta+s^2}}}}+\mathrm O(\epsilon^2)$$
This approximation is already very good - play with some concrete numerical examples here: https://www.desmos.com/calculator/fi3qefgyoe
