integral of sqrt[abs[x]]
Answer: \int_0^x \sqrt{|X|}dX = \frac{2}{3}x\sqrt{|x|}
Proof: Consider that for x\geq 0, we have
\int_0^x \sqrt{|X|}dX =\int_0^x \sqrt{X}dX = \frac{2}{3}x\sqrt{x},\mbox{ for }x\geq 0.
Also, if x\leq 0,, set t=-x so that t\geq 0, and we have
\int_0^x \sqrt{|X|}dX =\int_0^{-t} \sqrt{|X|}dX
now let u=-X so that du=-dX and 0\leq X\leq -t becomes 0\leq u\leq t and the integral becomes
\int_0^{-t} \sqrt{|X|}dX = -\int_0^{t} \sqrt{|-u|}du = -\int_0^{t} \sqrt{u}du = -\frac{2}{3}t\sqrt{t}= \frac{2}{3}x\sqrt{-x},\mbox{ for }x\leq 0
putting these togeather we have
\int_0^x \sqrt{|X|}dX =\left\{\begin{array}{cc}\frac{2}{3}x\sqrt{-x}, & \mbox{ if } x\leq 0\\ \frac{2}{3}x\sqrt{x},&\mbox{ if }<br />
x\geq 0\end{array}\right. =\frac{2}{3}x\sqrt{|x|}