I know that I can't integrate some unlimited functions while my interval of integration contains the singularity point. For example, $\int_{-1}^{1}\frac{1}{x}dx$ is undefined, it makes sense to me because $\int_{-1}^{0}\frac{1}{x} dx + \int_{0}^{1}\frac{1}{x} dx$ would be indeterminate (maybe this is wrong intuition) and doesn't matter if it is defined in $x=0$. However, there appeared a function: $$\frac{1}{\sqrt{\mid x \mid}}$$
And if there is a need, I can define an image at $x=0$. And I was in doubt if $$\int_{-1}^{1}\frac{1}{\sqrt{\mid x \mid}} dx = \int_{-1}^{0} \frac{1}{\sqrt{\mid x \mid}} dx + \int_{0}^{1}\frac{1}{\sqrt{\mid x \mid}} dx$$ is well defined integral since I can write it by the sum of two finite integrals. It might be a dumb question.
Thank you in advance.
By definition of improper integral $\displaystyle{\int_0^1} \dfrac{1}{\sqrt{x}}\ dx = \lim_{x \to 0} \int_x^1 \dfrac{1}{\sqrt{x}}\ dx = \lim_{x\to 0} (2\sqrt{1} - 2\sqrt{x}) = 2$ ($2\sqrt{x}$ is an antiderivative of $\dfrac1{\sqrt{x}}$). $\displaystyle{\int_{-1}^0}\dfrac1{\sqrt{-x}}\ dx = 2$ as well. Thus, $\displaystyle\int_{-1}^1 \dfrac1{\sqrt{|x|}}\ dx = 4.$