$\int_0^{2 \pi} \sinh^{-1} \frac{h}{\sqrt{s^2 + R^2 -2sR \cos(\phi)}} \, d\phi $

Question

The definite integral is the following one:

$$I_0 = \int_0^{2 \pi} \sinh^{-1} \frac{h}{\sqrt{s^2 + R^2 -2sR \cos(\phi)}} \, d \phi $$

Where $h, s, R$ are positive real quantities and $\sinh^{-1}$ is the arcsinh function. Basically, I have a cylinder of height $h$ and radius $R$ centered at the $xy$ plane origin. It goes up along the $z$ axis. I want to calculate the magnetic potential at a radial distance $s$ from the $z$ axis. And $\phi$ is the argument in respect to the $x$ axis. In other words, I am using the usual cylindrical coordinates, $s, \phi, z$.

This integral appeared when I was calculating the magnetic vector potential $A$. I tried to make substitutions with inverse functions ($\phi = \sin(x)$ and so on...) and also tried to substitute the whole argument of arcsinh so as to achieve an easier one. I also tried to use the approximation $\ln(2x) = \sinh^{-1}(x)$. But nothing worked out well. Also tried to expand in a power series the integrand itself and just the cosine function...

One of the results I found was:

$$I_0 = 4h^2 \int_{\sinh^{-1}(\frac{h}{s-R})}^{\sinh^{-1}(\frac{h}{s+R})} \frac{y \coth(y)}{\sqrt{-(\sinh^4(y)(s^2-R^2)^2-2(s^2+R^2)h^2 \sinh^2(y) +h^4 )}}dy$$

It appears to be a binomial expansion, but it isn't. It then becomes:

$$I_0 = 4h^2 \int_{\sinh^{-1}(\frac{h}{s-R})}^{\sinh^{-1}(\frac{h}{s+R})} \frac{y \coth(y)}{\sqrt{ (\sinh^2(y) 2sR)^2 - (\sinh^2(y)(s^2-R^2)-h^2)^2 }}dy$$

Any suggestions would be very much appreciated! Thanks!

MAJOR EDIT

It seems very unlikely that this integral above is easily solvable. That made me go back and realize the following:

$A(s, 0, 0) = \underbrace{\frac{u_0 nI h R}{4 \pi}}_{C} \displaystyle\int_0^{2 \pi} \displaystyle\int_{-1}^1 \frac{ \bar{\phi}}{\sqrt{s^2-2sR \cos(\phi) +R^2+h^2z^2}} dz d \phi$,

where $\bar{\phi}$ is the unit vector (cylindrical coordinates). Notice that, if we solve the integration in respect to $z$ we will achieve the nasty integral above ($I_0$). However, let's approach differently this time. Let's invert the order of integration and perform the following substitution: $\phi = \arccos(x)$. Then it becomes:

$A(s, 0, 0) = 2C \displaystyle\int_{-1}^1 \displaystyle\int_{-1}^1 \frac{\bar{\phi}}{ \sqrt{1-x^2} \cdot \sqrt{(s^2+R^2+h^2z^2)-2sRx}} dxdz$

Let's take both radicands to be non-negative and we then have:

$A(s, 0, 0) = 2C \displaystyle\int_{-1}^1 \displaystyle\int_{-1}^1 \frac{\bar{\phi}}{ \sqrt{ 2sRx^3 - (s^2+R^2+h^2z^2)x^2 - 2sRx + (s^2+R^2+h^2z^2) } } dxdz$

Which seems to be an elliptic function (if my quick search was correct). If these steps are correct, how should I proceed? Maybe residue theorem? Although I could recognize, I don't have the knowledge to solve it properly.

So far, I have found the following websites:

http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf ; https://functions.wolfram.com/EllipticIntegrals/EllipticE/introductions/CompleteEllipticIntegrals/ShowAll.html ; https://en.wikipedia.org/wiki/Elliptic_integral

Although they mentioned a similar integral to $A(s, 0, 0)$ and the need of elliptic integrals, they do not explain much.

Answer 1

You said numerical methods is an option so I'm only posting this because of that. This just uses the generic quad method which I think is Gauss-Legendre integration.

from scipy.integrate import quad
from numpy import sqrt,arcsinh, pi,cos

def integrand(x, h, s,R):

    return arcsinh(h/sqrt(s**2 +R**2 -2*s*R*cos(x)))
### parameters
h = 1
s = 1
R = 10

I = quad(integrand, 0, 2*pi,args=(h,s,R))

the results are given as a tuple. For instance this $(0.6288320765849464, 3.145647904875375e-09)$ means it approximates it as 0.6288 whatever within $3.25e-9$. You need to be careful about the constants because it doesn't evaluate well in some instances. You could add in something to catch negative parameters if you want

Answer 2

I think that we can make approximations.

First of all, we have $$\int_0^{2\pi} \sinh^{-1}(\text{argument})\,d\phi=2\int_0^{\pi} \sinh^{-1}(\text{argument})\,d\phi$$ What I think is that we could expand as (at least) two series : one around $\phi=0$ to compute the integral between $0$ and $\frac \pi 2$ and another one around $\phi=\frac \pi 2$ to compute the integral between $\frac \pi 2$ and $\pi$.

Using the same values as in @Ryan Howe's answer and using series to $O\left(t^{n+1}\right)$ and $O\left(\left(t-\frac{\pi }{2}\right)^{n+1}\right)$, I computed using these series $$I_1=2\int_0^{\frac\pi 2} \sinh^{-1}(\text{argument})\,d\phi\qquad I_2=2\int_{\frac\pi 2}^\pi \sinh^{-1}(\text{argument})\,d\phi$$

$$\left( \begin{array}{cccc} n & I_1 & I_2 & I_1+I_2 \\ 3 & 0.33073804 & 0.29589435 & 0.62663239 \\ 4 & 0.33531563 & 0.29420164 & 0.62951727 \\ 5 & 0.33531563 & 0.29435349 & 0.62966912 \\ 6 & 0.33404741 & 0.29461198 & 0.62865939 \\ 7 & 0.33404741 & 0.29444469 & 0.62849210 \\ 8 & 0.33445144 & 0.29447519 & 0.62892663 \\ 9 & 0.33445144 & 0.29449613 & 0.62894757 \\ 10 & 0.33431268 & 0.29447737 & 0.62879005 \\ 11 & 0.33431268 & 0.29448269 & 0.62879537 \\ 12 & 0.33436278 & 0.29448433 & 0.62884711 \end{array} \right)$$ while the "exact" values are $I_1=0.33434920$, $I_2=0.29448286$ and $I_1+I_2=0.62883206$.