This is just to clarify my doubt regarding absolute values functions.
Lets say there is a function $$f(x) = ax^{2} - \left|\frac{bx}{c}\right|$$
and we are asked to integrate this over $-\infty \to \infty$
$$\int_{-\infty}^{\infty} f(x) dx$$
So do we have to first redefine the function like this? $$f(x) = \begin{cases} ax^{2} - \left(\frac{bx}{c}\right), & x \ge 0 \\ ax^{2} + \left(\frac{bx}{c}\right), & x \lt 0 \end{cases}$$
and then redefine the integral like this?
$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} ax^{2} + \left(\frac{bx}{c}\right) dx + \int_{0}^{\infty} ax^{2} - \left(\frac{bx}{c}\right) dx$$
and what happens at 0? is this continuous at x = 0
Yes, you have the right idea. Notice that $$f(x)=ax^2-\left| \frac{bx}{c}\right|=ax^2-\left| \frac{b}{c}\right|\cdot |x|$$ And for the time being we will just set $j=\left| \frac{b}{c}\right|$. And what we know is that, now like you said, we have this piece wise function $$f(x) = \begin{cases} ax^{2} - jx, & x \ge 0 \\ ax^{2} + jx, & x \lt 0 \end{cases}$$ And therefore we have, like you said $$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} (ax^{2} + jx) dx + \int_{0}^{\infty} (ax^{2} - jx) dx$$ And notice that we have no issues with zero because the whole $jx$ term becomes zero, making the absolute value irrelevant. And yes, this function is continuous at $x=0$.
Also note, that since we can break $\left|\frac{bx}{c}\right|$ into $\left|\frac{b}{c}\right|\cdot |x|$, the only thing we have to worry about is $|x|$, which we did handle. This is because $\left|\frac{b}{c}\right|$ only amounts to a constant.