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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?

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

2 Answers 2

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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.

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  • $\begingroup$ The trick is to be considered dirty because $\sqrt{-1}$ is hybris. $\endgroup$
    – Gyro Gearloose
    Nov 28, 2015 at 17:42
  • $\begingroup$ Thank you. I understand, but I am a bit bewildered by this trick. It seems too wild. $\endgroup$
    – Emilio Novati
    Nov 28, 2015 at 18:00
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    $\begingroup$ "hybris", as in offending the gods by thinking one is their equal, or did you maybe mean "hybrid"? ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Nov 28, 2015 at 18:01
  • $\begingroup$ 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. $\endgroup$
    – imranfat
    Nov 28, 2015 at 18:03
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    $\begingroup$ 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). $\endgroup$
    – Rafael_Espericueta
    Nov 28, 2015 at 20:16
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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 $0$.

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  • $\begingroup$ +1 for pointing out that 2 constants are needed for a general solution. $\endgroup$
    – Gyro Gearloose
    Dec 1, 2015 at 11:08

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