convergence of series involving Appell function My aim (and hope) is to find a closed form the following integral
$$
I(z ; v, \delta v) = \int^z_0 \text{d}u \, \sqrt{z^2 - u^2} 
\exp\left[-\frac{(v-u)^2}{2 \,\delta v^2}\right]
$$
I tried what suggested in Integral: Gaussian times a square root, that is to say  


*

*integrating $I'(z;v, \delta v)$ and then integrating $I'$: 
didn't simplify the calculation   

*integrating $I^2(z;v, \delta v)$ in polar coordinates: same as above  

*use series expansion of $\exp$
I found the last one to be promising. Here's what I did:  


*

*changed variable from $u$ to $w = u-v$, giving 
$$
\int^{z-v}_{-v} \text{d}w \, \sqrt{z^2 - (w+v)^2} 
\exp\left[-\frac{w^2}{2 \,\delta v^2}\right]
$$

*used the series expansion of $e^{-x^2}$, giving . 
$$
\sum^\infty_{k=0} \left(-\frac{1}{2}\right)^k \frac{1}{k!(\delta v)^{2k}} 
\int^{z-v}_{-v} \text{d}w \,  w^{2k} \, \sqrt{z^2 - (w+v)^2} 
$$

*I calculated the integral with Mathematica. After some algebra, we get
$$ v\,\sqrt{z^2-v^2} \sum^\infty_{k=0} \left(-\frac{1}{2}\right)^k 
\left(\frac{v}{\delta v}\right)^{2k} \frac{1}{k!}
F_1\left(1+2k,-\frac{1}{2},-\frac{1}{2},2+2k,\frac{v}{v-z},
\frac{v}{v+z}\right)
$$  

*using property https://dlmf.nist.gov/16.16.E8 of the Appell function, this simplifies to
$$ v\, z \sum^\infty_{k=0} \left(-\frac{1}{2}\right)^k 
\left(\frac{v}{\delta v}\right)^{2k} \frac{1}{k!}
\frac{F_1\left(1,-\frac{1}{2},-\frac{1}{2},2+2k,\frac{v}{z},
-\frac{v}{z}\right)}{1+2k}
$$  

*I even tried expressing the Appell function into its integral representation, which gives
$$ v\, z \sum^\infty_{k=0} \left(-\frac{1}{2}\right)^k 
\left(\frac{v}{\delta v}\right)^{2k} \frac{1}{k!}
\frac{\Gamma(2+2k)}{(1+2k)\Gamma(1+2k)} 
\int^1_0 \text{d}u \, (1-u)^{2+2k} \sqrt{1-\frac{u^2v^2}{z^2}}
$$  

*using the properties of the $\Gamma$ function, this can be written as
$$ v \sum^\infty_{k=0} \left(-\frac{1}{2}\right)^k 
\left(\frac{v}{\delta v}\right)^{2k} \frac{1}{k!} 
\int^1_0 \text{d}u \, (1-u)^{2+2k} \sqrt{z^2-u^2v^2}
$$
This last passage looks harmless but I can't find a solution. I tried 
feeding both this and the expression at the fourth point to Mathematica, hoping to find the sum of the series.  
Does anyone have suggestions on how to proceed or can advice a way to 
find the sum of the series?  
I need this as $I(z ; v, \delta v)$ is itself an integrand of a more complex integral. I have also tried swapping integration order, in the hope that the integration in $z$ would simplfy the problem, but no luck on that front either. Any advice is highly appreciated.
 A: It's best to reduce the number of parameters (also I changed the notation $\delta v \to \delta$ to avoid confusion).
Substitute:
$$u= z t$$
$$I(z ; v, \delta) = z^2 \int^1_0 \text{d}t \, \sqrt{1 - t^2} 
\exp\left[-\frac{(v/z-t)^2}{2 \,\delta^2/z^2}\right]$$
Denoting:
$$\alpha = v/z, \qquad \beta=\delta/z$$
Now we only need to find:
$$J(\alpha, \beta)=\int^1_0 \text{d}t \, \sqrt{1 - t^2} 
\exp\left[-\frac{(\alpha-t)^2}{2 \beta^2}\right]$$
Trigonometric substitution might help, but then again, it might not.
1) We can expand the square root:
$$\sqrt{1 - t^2}=\sum_{n=0}^\infty \binom{2n}{n} \frac{t^{2n}}{(1-2n)4^n}$$
Which gives us:
$$J(\alpha, \beta)=\sum_{n=0}^\infty \binom{2n}{n} \frac{1}{(1-2n)4^n}\int^1_0   t^{2n} 
\exp\left[-\frac{(t-\alpha)^2}{2 \beta^2}\right] \text{d}t$$
The integrals are not that simple, however they can be expressed in terms of error function.
$$\int^1_0   t^{2n} 
\exp\left[-\frac{(t-\alpha)^2}{2 \beta^2}\right] \text{d}t= \\ = \int^{1-\alpha}_{-\alpha}   (t+\alpha)^{2n} 
\exp\left[-\frac{t^2}{2 \beta^2}\right] \text{d}t= \\ = \sqrt{2} \beta \int^{(1-\alpha)/(\sqrt{2} \beta)}_{-\alpha/(\sqrt{2} \beta)}   (\sqrt{2} \beta t+\alpha)^{2n} 
e^{-t^2} \text{d}t$$
Yeah, this is not pretty in general.
2) We can expand the exponential, as the OP did:
$$J(\alpha,\beta)=\sum_{n=0}^\infty \frac{(-1)^n \alpha^{2 n}}{n! 2^n \beta^{2n}} \int_0^1  \sqrt{1 - t^2} \left(1- \frac{1}{\alpha} t \right)^{2 n} \text{d}t$$
There's no need to involve Appell function here, we can use the binomial theorem:
$$\int_0^1  \sqrt{1 - t^2} \left(1- \frac{1}{\alpha} t \right)^{2 n} \text{d}t=\sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^k}{\alpha^k} \int_0^1  \sqrt{1 - t^2} ~t^k dt$$
$$\int_0^1  \sqrt{1 - t^2} ~t^k dt=  \\ =\frac{1}{2} \int_0^1  \sqrt{1 - t} ~t^{(k-1)/2} dt= \frac{1}{2} B \left(\frac{k+1}{2}, \frac{3}{2} \right)= \\ = \frac{\sqrt{\pi}}{4} \frac{\Gamma \left(\frac{k+1}{2}\right)}{\Gamma \left(\frac{k}{2}+2 \right)}$$
So we have:
$$J(\alpha,\beta)=\frac{\sqrt{\pi}}{4} \sum_{n=0}^\infty \frac{(-1)^n \alpha^{2 n}}{n! 2^n \beta^{2n}} \sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^k}{\alpha^k} \frac{\Gamma \left(\frac{k+1}{2}\right)}{\Gamma \left(\frac{k}{2}+2 \right)}$$
The inner sum is clearly hypergeometric, but we need to separate even and odd terms to make any sense of it.
