Compute $\int_{a-b}^{a+b} \chi_{(-t,t)}(y)dt$

Question

Compute $\int_{a-b}^{a+b} \chi_{(-t,t)}(y)dt$.

So if I create a number line marking a-b and a+b. If that the integral above has 5 different answers depending on where (-t,t) is located on the number line.

• Of course if (-t,t) if before the mark of a-b and a+b on the number line then the integral is equal to zero.
• If a bit of (-t,t) hangs outside of the mark of a-b but some of it is in (a-b,a+b), then the integral is equal to y-a+b.
• Similarly, if a bit of (-t,t) hangs outside of the mark of a+b but some of it is in (a-b,a+b), then the integral is equal to a+b+t-y.
• If (-t,t) is contained in (a-b,a+b) then the integral is equal to 0.

So what I have is

$$\int_{a-b}^{a+b} \chi_{(-t,t)}(y)dt=\begin{cases} 0, & \text{if }y<a-b \\ y-a+b, & \text{if } a-b\leq y \leq ? \\ 1, & \text{if }?\leq y \leq ?\\ a+b+t-y, & \text{if }?\leq y \leq a+b \\ 0, & \text{if }y > a+b \\ \end{cases}$$

What would go in the "?" area?

Answer 1

I'm assuming $\chi_{(-t,t)}$ is the characteristic function of $(-t,t)$, i.e. \begin{equation} \chi_{(-t,t)}(y) = \begin{cases} 1, & y \in (-t,t), \\ 0, & y \notin (-t,t). \end{cases} \end{equation} The way you explain the cases isn't clear, though, as you state that they depend on $t$? But $t$ is the variable of integration, it ranges from $a-b$ to $a+b$, so the cases should be in terms of $y$, $a$, and $b$ (consistent with the incomplete answer you give).

Also, regarding the first case stated: if $y < a-b$, then it is still possible that $y \in (-t,t)$ for $t \in (a-b,a+b)$ (so the integral is nonzero). For instance, if $a=6$, $b=1$, and $y=-3$, then we have $\int_5^7 \chi_{(-t,t)}(-3) dt = \int_5^7 1 dt = 2$, as $-3\in (-t,t)\;\forall t \in (5,7)$. So I disagree with the first case listed.

Instead, I think there are a lot of cases to consider based on the signs of $a$ and $b$. Let me take the liberty of assuming $0 \leq b \leq a$ so that $0 \leq a-b \leq a+b$ for simplicity. $$ \longleftarrow \!\!\!-\!\!\!-\!\!\!\!\!\!\! \stackrel{-(a+b)}{\bullet} \!\!\!\!\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\!\!\!\! \stackrel{-(a-b)}{\bullet} \!\!\!\!\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\! \stackrel{0}{\bullet} \!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\! \stackrel{a-b}{\bullet} \!\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\! \stackrel{a+b}{\bullet} \!\!-\!\!\!-\!\!\!\!\!\!\!\!\! \longrightarrow $$ Then I claim \begin{equation} \int\limits_{a-b}^{a+b}\chi_{(-t,t)}(y)dt = \begin{cases} 0, &y \leq -(a+b),\\ a+b+y, &-(a+b) < y \leq -(a-b),\\ 2b, &-(a-b) < y \leq a-b,\\ y-(a-b), &a-b < y \leq a+b,\\ 0, &a+b < y. \end{cases} \end{equation} To justify this: The integrand is always $0$ or $1$, so we just compute the size of the interval $I = \{t\in(a-b,a+b):y\in(-t,t)\}$.

If $y \leq -(a+b)$, then $y \notin (-t,t)$ whenever $a-b<t<a+b$, so we get $I = \emptyset$.

If $-(a+b)<y\leq-(a-b)$, then $I$ is the interval from $-(a+b)$ to $y$.

If $-(a-b) < y \leq a-b$, then $y\in(-t,t)$ for every $t$ under consideration, so $I = (a-b,a+b)$.

etc.

However, we need the assumption $0 \leq b \leq a$. For instance, if $a$ and $b$ can be anything, then the first condition becomes \begin{equation} \int\limits_{a-b}^{a+b}\chi_{(-t,t)}(y)dt = \begin{cases} 0, &y \leq \min\{\pm(a-b),\pm(a+b)\},\\ \vdots \end{cases} \end{equation}

Answer 2

Actually, you can change it a little, $\chi_{(-t,t)} (y)=\chi_{(\max(-y,y), +\infty)} (t)=\chi_{(|y|,+\infty)} (t),$ so we change the integral to $J=\int_{a-b}^{a+b} \chi_{(|y|, +\infty)} (t) dt= \mu((|y|, +\infty)\cap (a-b, a+b ))$.

When $|y| \leq a-b, J=2b.$

When $|y| \geq a+b, J=0.$

When $a-b<|y|<a+b, J=a+b-|y|$.

Then we consider the sign of $y$ and the $b-a\geq 0$ or not.