Analytic formula for $\int_{-\pi}^\pi h(\cos\theta)h(t\cos\theta +\sqrt{1-t^2}\sin\theta)d\theta$, where $h(t) := \mathrm{sign}(t)$

Question

For a continuous funciton $h:[-1,1] \to \mathbb R$, consider the function $u_h:[-1,1] \to \mathbb R$ defined by $$ u_h(t) := \int_{-\pi}^\pi h(\cos\theta)h(t\cos\theta +\sqrt{1-t^2}\sin\theta)d\theta. $$

If $h_1$ is the function defined by $h_1(t) = 1$ if $t \ge 0$ and $h(t) = 0$ otherwise, then, noting that for $\theta \in [-\pi,\pi]$, we have $\cos\theta \ge 0$ iff $\theta \in [-\pi/2,\pi/2]$, one computes

$$ u_{h_1}(t) = \int_{-\pi/2}^{\pi/2} h_1(t\cos\theta +\sqrt{1-t^2}\sin\theta)d\theta. $$ Now, for $\theta \in [-\pi/2,\pi/2]$, one has $$ \begin{split} t\cos\theta +\sqrt{1-t^2}\sin\theta \ge 0 &\iff \tan(-\theta) \le \frac{t}{\sqrt{1-t^2}} = \tan(\arcsin(t))\\ &\iff \theta \ge -\arcsin(t). \end{split} $$ Thus, we obtain the following analytic formula $$ u_{h_1}(t) = \int_{-\arcsin(t)}^{\pi/2}d\theta=\arcsin(t) + \frac{\pi}{2}=\arccos(-t). $$

---

Now consider the function $h_2(t) := \mathrm{sign}(t) = t / |t|$ with $h_2(0) := 0$. We wish to compute $u_{h_2}$. Unfortunately, playing a game similar to the above leads to an explosion of subcases to consider.

Question. What is an analytic formula for $u_{h_2}(t)$ for all $t \in [-1,1]$ ?

Observation. $h_2(t) = h_1(t)-h_1(-t)$ for all $t \in [-1, 1]$.

Answer 1

Repeating and following through on my comment:

$h[\cos(\theta)]~$ is $~+1~$ if $~\cos(\theta)> 0~$, and is $~−1~$ if $\cos(\theta)<0.$

$t$ is some unknown fixed element in $[−1,1].$

Let fixed $\alpha$ be chosen from $[−\pi/2,\pi/2]$ so that
$\sin(\alpha) = t~$, which implies that $~\cos(\alpha) = \sqrt{1 − t^2},~$ which is always non-negative.

Then, the original integral becomes

$$\int_{-\pi}^{\pi} ~h\cos(\theta) ~~h[\sin(\alpha + \theta)] ~dθ.\tag1$$

I am going to confine my answer to the presumption that $0 < \alpha < \pi/2.$
The analysis when $-\pi/2 < \alpha < 0$ will be very similar.

To evaluate the integral in (1) above, you must analyze what happens when $\theta$ ranges from $-\pi$ to $\pi$.

The question is, in what proportion of this interval, which has a width of $(2\pi)$, will $~h\cos(\theta)~$ and $~h[\sin(\alpha + \theta)]~$ be either both positive or both negative, rather than one positive and one negative.

---

$h[\cos(\theta)]$ will be positive if

• $-\pi/2 < \theta < \pi/2.$

$h[\cos(\theta)]$ will be negative if either

• $-\pi < \theta < -\pi/2~$ or
• $~\pi/2 < \theta < \pi.$

---

$h[\sin(\alpha + \theta)]$ will be positive if

• $-\alpha < \theta < \pi - \alpha.$

$h[\sin(\alpha + \theta)]$ will be negative if either

• $-\pi < \theta < -\alpha~$ or
• $\pi - \alpha < \theta < \pi.$

---

$h[\cos(\theta)]~$ and $~h[\sin(\alpha + \theta)]~$ will both be positive if

• $-\alpha < \theta < \pi/2.$
  Interval width $= \pi/2 + \alpha$.

$h[\cos(\theta)]~$ and $~h[\sin(\alpha + \theta)]~$ will both be negative if

• $-\pi < \theta < -\pi/2.$
  Interval width $= \pi/2$.

Thus, the combined interval in which $h[\cos(\theta)]~$ and $~h[\sin(\alpha + \theta)]~$ will have the same sign has width $\pi/2 + \alpha + \pi/2 = \pi + \alpha.$

Therefore, since the overall interval has width $(2\pi)$, the combined interval in which $h[\cos(\theta)]~$ and $~h[\sin(\alpha + \theta)]~$ will have different signs must have width $\pi - \alpha.$

Therefore, the integral in (1) above must evaluate to

$$ (\pi + \alpha) - (\pi - \alpha) = 2\alpha.$$