# A ''strange'' integral from WolframAlpha

I want integrate: $$\int \frac{1}{\sqrt{|x|}} \, dx$$ so I divide for two cases $$x>0 \Rightarrow \int \frac{1}{\sqrt{x}} \, dx= 2\sqrt{x}+c$$

$$x<0 \Rightarrow \int \frac{1}{\sqrt{-x}} \, dx= -2\sqrt{-x}+c$$ But WolframAlpha gives: $$\int \frac{1}{\sqrt{|x|}} \, dx=\left(\sqrt{-x}+\sqrt{x} \right)\operatorname{sgn}(x)-\sqrt{-x}+\sqrt{x} +c$$ How I can interpret this result? Maybe I'm wrong?

• Wolfram Alpha gives $\left(\sqrt{-x}+\sqrt{x} \right)\mbox{sgn}(x)-\sqrt{-x}\color{red}{+}\sqrt{x}+c$ Nov 28, 2015 at 17:22
• In your results, you wrote $\dfrac{1}{2}$ instead of $2$. Nov 28, 2015 at 17:22

Using $\operatorname{sgn}(x)$ is just a (half-dirty) trick to put the two cases into one. Put in $-1$ vs. $+1$ for $\operatorname{sgn}(x)$ and your eyes will be open.

• The trick is to be considered dirty because $\sqrt{-1}$ is hybris. Nov 28, 2015 at 17:42
• Thank you. I understand, but I am a bit bewildered by this trick. It seems too wild. Nov 28, 2015 at 18:00
• "hybris", as in offending the gods by thinking one is their equal, or did you maybe mean "hybrid"? ${}\qquad{}$ Nov 28, 2015 at 18:01
• I think this is a good example (+1) that we have to be cautious what online tools provide us as an answer. Not that it is wrong, but certainly an eye opener and it is justified to be a bit critical. Nov 28, 2015 at 18:03
• It's just a way to write it in one line without having to break it down into two separate cases (this sort of thing is frequently done). Nov 28, 2015 at 20:16

Since the function is not defined for $x=0$, it's not really meaningful to have a single constant of integration for the whole thing. The most general function $F$ (not defined at $0$) for which, at each point $x\ne0$, $F'(x)=\frac{1}{\sqrt{|x|}}$, is $$F(x)=\begin{cases} -2\sqrt{-x}+c_1 & \text{if x<0}\\ 2\sqrt{x}+c_2 & \text{if x>0} \end{cases}$$ where $c_1$ and $c_2$ are arbitrary constants.

Among these functions there are some that can be extended by continuity at $0$, namely those for which $c_1=c_2$, but they're just a special case. Note that none of these special functions is differentiable at $0$.

• +1 for pointing out that 2 constants are needed for a general solution. Dec 1, 2015 at 11:08