# 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.

• I think you should expand everything. Then you basically have three integrals:$$C_1\int d\phi'\cos^2\frac{\phi'}2,\quad C_2\int d\phi'\cos^3\phi'\cos^2\frac{\phi'}2,\quad\text{and}\quad C_3\int d\phi'\cos 2\phi'\cos^2\frac{\phi'}2,$$where $C_1,C_2,C_3$ are constants (with respect to $\phi'$). – Friedrich Philipp Mar 18 '16 at 14:40
• But my problem is that, even if I do so, the outer integral is not solvable anymore. For example, $$C_1\int d\phi'\cos^2\frac{\phi'}2=\frac{1}{2} C_1\left(\cos ^{-1}(\Delta/V+\cos \phi)+\sqrt{1-(\Delta/V+\cos \phi)^2}\right),$$ which, I think, is not integrable with respect to $\phi'$. Maybe you can suggest a good approximation? – Elina Mar 18 '16 at 14:53
• I did not say that it gets simpler. ;o) But this is integrable over $[0,\pi]$. Note that $C_1$ also highly depends on $\phi$. But honestly, I don't envy you. ;o) – Friedrich Philipp Mar 18 '16 at 15:02
• @Elina I believe this integral may be solved exactly in terms of elliptic integrals in the case you are interested in. Looks tough but certainly doable. Out of curiosity, what exactly is the physical context of the problem you are trying to solve? – David H Mar 20 '16 at 3:55
• As far as I understand, the integrand of an elliptic integral needs to be a rational function, but I have arccos here... @DavidH, it's electrons scattering off a magnetic impurity. I have an incoming electron with an incidence angle $\phi$ and energy $\varepsilon$, and I'm calculating the probability that it will scatter at an angle $\phi'$ and energy $\varepsilon + \Delta$, and by doing so flip the spin of the magnetic impurity. – Elina Mar 21 '16 at 12:38

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!

• Thank you for the help! It all boils down to these two integrals (the rest are doable): $$\int\limits_{-1}^{1-a}dt\,\frac{(1-t)\sqrt{1-(t+a)^2}(18t^3+(30a-8)t^2+(6a^2+2a-9)t-6a^3+10a^2+15a-10)}{72\sqrt{1-t^2}}$$ and $$\int\limits_{-1}^{1-a}dt\,\frac{(1-t)\arcsin(t+a)(8t^3+12at^2-4a^3+6a-3)}{24 \sqrt {1-t^2}}.$$ Do you know how to handle this? I looked at the generalized hypergeometric functions, but didn't right away see how to use them here. – Elina Mar 22 '16 at 18:37
• @Elina Sorry for taking so long getting back to you. Of the two integrals you mentioned, the first is of course an elliptic integral, making this integral in principle quite easy, though computing it by hand would involve tortuous amounts of algebra. The second integral is much more subtle, and belongs to a family of integrals I've been studying off and on for nearly a year. See my question here: math.stackexchange.com/questions/1300459/… . There I find a solution in terms of Appell $F_3$ functions. – David H Mar 31 '16 at 3:32
• @Elina I would also like to impress upon that I spent months searching in vain for a simpler solution. First I tried to reduce my Appell $F_3$ function to elliptic integrals and gave up. Then I tried to reduce it to single-variable (generalized) hypergeometric functions and gave up. Finally I tried to reduce it to Appell $F_1$ or $F_2$ functions (which are somewhat simpler than the $F_3$ functions, relatively speaking), and again I gave up. But while the general case doesn't seem simplify, some particular cases nevertheless do. It would be very interesting if your integral is one of these. – David H Mar 31 '16 at 3:48