Assume random variable $$X \sim f_X(x) \,\,\, -2 \leq x\leq 2$$
Now Assume we need to compute the following
$$F= \mathbb{E}\left(\frac{1}{1+(G(X))^2}\right)$$ where we define the function $$G(x) = \mathbb{1}(|x|\le c) = \begin{cases} 1, & -c \le x \le c \\ 0, & \text{otherwise} \end{cases}$$
with $c<2$...
Can I say that the following is correct
$$F = \int_{-2}^{2}\left(\frac{1}{1+(G(x))^2}\right) f_X(x)\, dx =\dfrac{1}{2}\int_{-c}^{c} f_X(x)\, dx+ \int_{-2}^{-c} f_X(x)\,dx\,+\int_{c}^{2} f_X(x)\,dx$$
I can varify that $F$ is correctly calculated based on given definitions in the question. It can however be furher simplified as:
$$\dfrac{1}{2}\int_{-c}^{c} f_X(x)\, dx+ \int_{-2}^{-c} f_X(x)\,dx\,+\int_{c}^{2} f_X(x)\,dx=1-\dfrac{1}{2}\int_{-c}^{c} f_X(x)\, dx$$