3-dimensional integral with a step function I have a physics problem and it involves the following integral:
$$\int_0^{\pi} d\phi \int_0^{\pi} d\phi' \int_{V\cos\phi'-\Delta}^{ V\cos\phi} d\varepsilon \,
\varepsilon (\varepsilon+\Delta) \,
\theta\left(\cos\phi-\cos\phi'+\frac{\Delta}{V}\right)
\sin^2\frac{\phi}{2}\cos^2\frac{\phi'}{2}.$$
The step function $\theta(x)=1$ if $x>0$, and $\theta(x)=0$ otherwise. Both $V,\Delta\geq 0$.
If $\Delta/V=0$ or $\Delta/V>2$, then it's easy to get rid of the step function and solve the integral exactly. I'm interested in the case $0<\Delta/V<2$. In this case for the step function to be non-zero we need $\phi'<\arccos(\cos\phi+\Delta/V)$. After performing the integration over $\varepsilon$, we therefore get
$$\int\limits_{0}^{\pi} d\phi \int\limits_0^{\arccos(\cos\phi+\Delta/V)} d\phi' \,
\frac{1}{12} \left(V^3\left(3 \cos \phi+\cos 3 \phi-4 \cos^3\phi'\right) \\
+3V^2 \Delta (\cos 2\phi+\cos 2 \phi' +2)-2 \Delta^3\right)
\sin^2\frac{\phi}{2}\cos^2\frac{\phi'}{2}.$$
How should I proceed from here? It's possible to do the inner integral over $\phi'$, but the result is very lengthy and not integrable.
 A: HINT:
For $\Delta\in\mathbb{R}^{\ge0}\land V\in\mathbb{R}^{+}$, define $\Theta{\left(V,\Delta\right)}$ via the integral,

$$\small{\Theta{\left(V,\Delta\right)}:=\int_{0}^{\pi}\mathrm{d}\varphi\,\sin^{2}{\left(\frac{\varphi}{2}\right)}\int_{0}^{\pi}\mathrm{d}\varphi^{\prime}\,\cos^{2}{\left(\frac{\varphi^{\prime}}{2}\right)}\int_{V\cos{\left(\varphi^{\prime}\right)}-\Delta}^{V\cos{\left(\varphi\right)}}\mathrm{d}\epsilon\,\epsilon\left(\epsilon+\Delta\right)\,H{\left(\cos{\left(\varphi\right)}-\cos{\left(\varphi^{\prime}\right)}+\frac{\Delta}{V}\right)}}.$$

Assume $0<\Delta<2V$. Then, substituting $\cos{\left(\varphi\right)}=t$ and $\cos{\left(\varphi^{\prime}\right)}=u$, we find:
$$\begin{align}
\Theta{\left(V,\Delta\right)}
&=\small{\int_{0}^{\pi}\mathrm{d}\varphi\,\sin^{2}{\left(\frac{\varphi}{2}\right)}\int_{0}^{\pi}\mathrm{d}\varphi^{\prime}\,\cos^{2}{\left(\frac{\varphi^{\prime}}{2}\right)}\int_{V\cos{\left(\varphi^{\prime}\right)}-\Delta}^{V\cos{\left(\varphi\right)}}\mathrm{d}\epsilon\,\epsilon\left(\epsilon+\Delta\right)\,H{\left(\cos{\left(\varphi\right)}-\cos{\left(\varphi^{\prime}\right)}+\frac{\Delta}{V}\right)}}\\
&=\small{\int_{-1}^{1}\mathrm{d}t\,\frac{1-t}{2\sqrt{1-t^{2}}}\int_{-1}^{1}\mathrm{d}u\,\frac{1+u}{2\sqrt{1-u^{2}}}\int_{Vu-\Delta}^{Vt}\mathrm{d}\epsilon\,\epsilon\left(\epsilon+\Delta\right)\,H{\left(t-u+\frac{\Delta}{V}\right)}}\\
&=\small{\frac{V^{3}}{4}\int_{-1}^{1}\mathrm{d}t\,\frac{1-t}{\sqrt{1-t^{2}}}\int_{-1}^{1}\mathrm{d}u\,\frac{1+u}{\sqrt{1-u^{2}}}\int_{u-\frac{\Delta}{V}}^{t}\mathrm{d}v\,v\left(v+\frac{\Delta}{V}\right)\,H{\left(t-u+\frac{\Delta}{V}\right)}};~~~\small{\left[\frac{\epsilon}{V}=v\right]}.\\
\end{align}$$
Let $a$ denote the parameter $a:=\frac{\Delta}{V}$. Note that $\left(0<a<2\land-1< t<1\right)$ implies
$$1-a<t<1\iff0<t-1+a<t-u+a,$$
and
$$-1<t<1-a\implies\left(-1<u<t+a\iff0<t-u+a\right).$$
Thus,
$$\begin{align}
\Theta{\left(V,\Delta\right)}
&=\small{\frac{V^{3}}{4}\int_{-1}^{1}\mathrm{d}t\,\frac{1-t}{\sqrt{1-t^{2}}}\int_{-1}^{1}\mathrm{d}u\,\frac{1+u}{\sqrt{1-u^{2}}}\int_{u-a}^{t}\mathrm{d}v\,v\left(v+a\right)\,H{\left(t-u+a\right)}}\\
&=\frac{V^{3}}{4}\int_{1-a}^{1}\mathrm{d}t\,\frac{1-t}{\sqrt{1-t^{2}}}\int_{-1}^{1}\mathrm{d}u\,\frac{1+u}{\sqrt{1-u^{2}}}\int_{u-a}^{t}\mathrm{d}v\,v\left(v+a\right)\\
&~~~~~\small{+\frac{V^{3}}{4}\int_{-1}^{1-a}\mathrm{d}t\,\frac{1-t}{\sqrt{1-t^{2}}}\int_{-1}^{1}\mathrm{d}u\,\frac{1+u}{\sqrt{1-u^{2}}}\int_{u-a}^{t}\mathrm{d}v\,v\left(v+a\right)\,H{\left(t-u+a\right)}}\\
&=\frac{V^{3}}{4}\int_{1-a}^{1}\mathrm{d}t\,\frac{1-t}{\sqrt{1-t^{2}}}\int_{-1}^{1}\mathrm{d}u\,\frac{1+u}{\sqrt{1-u^{2}}}\int_{u-a}^{t}\mathrm{d}v\,v\left(v+a\right)\\
&~~~~~+\frac{V^{3}}{4}\int_{-1}^{1-a}\mathrm{d}t\,\frac{1-t}{\sqrt{1-t^{2}}}\int_{-1}^{t+a}\mathrm{d}u\,\frac{1+u}{\sqrt{1-u^{2}}}\int_{u-a}^{t}\mathrm{d}v\,v\left(v+a\right).\\
\end{align}$$
Now the required integrals have algebraic integrands and are also step-function free, making it more susceptible to the usual integration techniques.
Also, I should point out that my comment above that the integral is ultimately elliptic may have been too hasty. Unless there is fortuitous cancellations, the integration actually require generalized hypergeometric functions. Good luck!
