Integration with double absolute value

Question

How to prove that?

$$ \int\left|\sqrt{|x|}\right|\ \mathrm{d}x = \frac{1}{3} x \left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x) + \sqrt{x} + \sqrt{-x}\right) $$

I cannot understand this result. Where does the $1/3$ factor come out from?

Answer 1

I find the outer $| \,|$ incomprehensible since $\sqrt{|x|}\ge0$ according to the usual definition of square root. I will answer ignoring that.

For $x\ge 0$ we have the usual result $\int x^{1/2}\,dx={2\over 3}x^{3/2}$ plus constant, which you can verify by taking the derivative of ${2\over 3}x^{3/2}$.

Suppose $x\ge 0$; then $\mathrm{ sgn} (x)=1$, and

$$\frac{1}{3} x \left((\sqrt{x} - \sqrt{-x})\mathrm{sgn}(x) + \sqrt{x} + \sqrt{-x}\right)={1\over 3}x(2\sqrt{x})={2\over 3}x^{3/2}$$

as desired (but that raises the issue of what $\sqrt{-x}$ is, perhaps $i\sqrt{x}$).

Now suppose $x\le 0$. We then have $\mathrm{ sgn} (x)=-1$, and we get $$\frac{1}{3} x \left(\left(\sqrt{x} - \sqrt{-x}\right)\mathrm{sgn}(x) + \sqrt{x} + \sqrt{-x}\right)={1\over 3}x(2\sqrt{-x})=-{2\over 3}(-x)^{3/2}.$$ Differentiated this also gives the right answer, $\sqrt{-x}$, and thus our expression equals $\int |x|^{1/2}\,dx=\int (-x)^{1/2}\,dx$, for $x\le 0$.