Integral with theta function

Question

$$I=\int_{-L/2}^{L/2}dx \Theta(\epsilon-2A|x|)$$ $L>0; A>0$ and $\epsilon$ are parameters, I should solve this integral but I don't know how I thought about solving it like this: $$\epsilon-2A|x|>0$$ so that $$\epsilon>2A|x|$$ but from here on I don't know how to go on. These are the solutions of the integral reported in the book:

$I=\dfrac{2\pi mL\epsilon}{A}$ for $\epsilon\leq AL$

$I=2\pi mL^2$ for $\epsilon> AL$ where $h$ is Planck's constant.

The Heavyside $\Theta$ function is so defined:

$$ \Theta(x) = \begin{cases} 1 & \text{ if } x \geq 0\\ 0 & \text{if } x < 0\\ \end{cases} $$

I hope you can help me.

Answer 1

You have a good start. \begin{align} \theta(\epsilon-2A|x|) = 1 \quad&\Longleftrightarrow\quad \epsilon-2A|x| \ge 0 \\ &\Longleftrightarrow\quad \epsilon \ge 2A|x| \\ &\Longleftrightarrow\quad |x| \le \frac{\epsilon}{2A} \end{align} So, if $\frac{\epsilon}{2A} \ge \frac{L}{2}$, then $$ I = \int_{-L/2}^{L/2} 1\;dx = L $$ on the other hand, if $\frac{\epsilon}{2A} < \frac{L}{2}$, then $$ I = \int_{-\epsilon/(2A)}^{\epsilon/(2A)} 1\;dx = \frac{\epsilon}{A} $$ Of course, my condition $\frac{\epsilon}{2A} \ge \frac{L}{2}$ is the same as your condition $\epsilon \ge AL$.

Answer 2

Hint:

Split the integral into positive $x$ and negative $x$

$$I=\int_{-L/2}^{L/2}dx \theta(\epsilon-2A|x|)=\int_{-L/2}^0dx \theta(\epsilon-2A|x|) +\int_{0}^{L/2}dx \theta(\epsilon-2A|x|)$$ $$=\int_{-L/2}^0dx \theta(\epsilon+2Ax) +\int_{0}^{L/2}dx \theta(\epsilon-2Ax).$$

Now, split the integrals such that the Heavyside theta function is well separated.