Does the following integral converge or diverge and can it be solved? I have been stuck solving the following integral for some time now: $$\int_0^{2R} \frac{\sin^2\left(\frac{x}{4R}\pi\right)R^2}{\left(\sqrt{\left(d+Z\right)^2+\sin^2\left(\frac{x}{4R}\pi\right)}\right)^3}\,\mathrm{d}x$$
If it isn't possible to solve, is there a way to show whether this integral converges or diverges for $\lim_\limits{R\to\infty}$?
Any help or suggestions on how to solve this are appreciated.
 A: If you are only interested in convergence when $R\rightarrow\infty$, you can observe that
$$
0 \le K\sin^2(\frac{x}{4R} \pi)\le \frac{\sin^2(\frac{x}{4R} \pi)R^2}{\left(\sqrt{(d+Z)^2+\sin^2(\frac{x}{4R} \pi)}\right)^3} = I
$$
with
$$
K=\frac{R^2}{\left(\sqrt{(d+Z)^2+1)}\right)^3}
$$
By consequence
$$
\int_R^{2R}  K\sin^2(\frac{x}{4R} \pi)dx \le \int_0^{2R}  K\sin^2(\frac{x}{4R} \pi)dx \le I
$$
But for $x\in[R,2R]$, $\frac{1}{2}=\sin^2(\frac{\pi}{4})\le \sin^2(\frac{x}{4R} \pi)$
By consequence
$$
\frac{KR}{2}=\int_{R}^{2R}\frac{K}{2}dx \le \int_R^{2R}  K\sin^2(\frac{x}{4R} \pi)dx
$$
and your integral $\rightarrow \infty$ when $R\rightarrow \infty$

Note: If you want an analyitcal expression you will have to use elliptic integrals: https://en.wikipedia.org/wiki/Elliptic_integral
A: First note that your integral diverges for $d + Z = 0$. For $d + Z \neq 0$ we can write
\begin{align}
I (R,d,Z) &\equiv \int \limits_0^{2R} \frac{R^2 \sin^2 \left(\frac{\pi x}{4R}\right)}{\left[(d+Z)^2 + \sin^2 \left(\frac{\pi x}{4R}\right)\right]^{3/2}} \, \mathrm{d} x \stackrel{x = \frac{4 R t}{\pi}}{=} \frac{4 R^3}{\pi} \int \limits_0^{\pi/2} \frac{\sin^2(t)}{[(d+Z)^2 + \sin^2(t)]^{3/2}} \, \mathrm{d} t \\
&\equiv \frac{4 R^3}{\pi} f(\lvert d + Z \rvert) \, .
\end{align}
Here we have defined the function
$$ f \colon (0,\infty) \to (0,\infty) \, , \, f(a) = \int \limits_0^{\pi/2} \frac{\sin^2(t)}{[a^2 + \sin^2(t)]^{3/2}} \, \mathrm{d} t \, .$$
Since $f$ is always positive, we immediately see that
$$ \lim_{R \to \infty} I (R,d,Z) = \lim_{R \to \infty} \frac{4 R^3}{\pi} f(\lvert d + Z \rvert) = \infty \, .$$
In order to compute your integral we need to find $f$. This is somewhat tricky and involves elliptic integrals, but it can be done: For $a > 0$ we have
\begin{align}
f(a) &= \int \limits_0^{\pi/2} \frac{1 + a^2 - \cos^2(t) - a^2}{[1 + a^2 - \cos^2(t)]^{3/2}} \, \mathrm{d} t = \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 + a^2 - \cos^2(t)}} - \frac{a^2}{[1 + a^2 - \cos^2(t)]^{3/2}}\right] \mathrm{d} t \\
&= \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 + a^2} \sqrt{1 - \frac{\cos^2(t)}{1+a^2}}} - \frac{1}{(1+a^2)^{3/2}} \left(\frac{a^2}{\left(1 - \frac{\cos^2(t)}{1+a^2}\right)^{3/2}} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\sin(t) \cos(t)}{\sqrt{1 - \frac{\cos^2(t)}{1+a^2}}}\right)\right] \mathrm{d} t \\
&= \frac{1}{\sqrt{1+a^2}} \int \limits_0^{\pi/2} \left[\frac{1}{\sqrt{1 - \frac{\cos^2(t)}{1+a^2}}} - \sqrt{1 - \frac{\cos^2(t)}{1+a^2}}\right] \mathrm{d} t \\
&= \frac{\operatorname{K} \left(\frac{1}{\sqrt{1+a^2}}\right) - \operatorname{E} \left(\frac{1}{\sqrt{1+a^2}}\right)}{\sqrt{1+a^2}} \, ,
\end{align}
where $\operatorname{K}$ and $\operatorname{E}$ are the complete elliptic integrals of the first and second kind, respectively. Note that the integral of the derivative term in the second line vanishes, so we are allowed to add it without changing the result.
