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We are on the real domain.

We have that $\dfrac 1 {\sqrt {x^2 - a^2} }$ is defined only for $|x| > a$.

It is straightforward to evaluate $\displaystyle \int \dfrac {\mathrm d x} {\sqrt {x^2 - a^2} }$ for $x > a$.

It comes out to $\cosh^{-1} \dfrac x a$.

However, it's not so straightforward when $x < -a$.

Substitute $z = -x$ to put it into the positive real domain, and get $\mathrm d z = -\mathrm d x$.

Using integration by substitution you get:

$$\int \dfrac {\mathrm d x} {\sqrt {x^2 - a^2} } = -\int \dfrac {\mathrm d z} {\sqrt {z^2 - a^2} }$$

which seems to work out to $-\cosh^{-1} \left({\dfrac z a}\right) = -\cosh^{-1} \left({\dfrac {-x} a}\right) = -\cosh^{-1} \left|{\dfrac x a}\right|$.

But I don't believe that, I think that minus sign needs to go, because I find in the book (R.P. Gillespie's "Integration") that he sort of expects it to be $\cosh^{-1} \left|{\dfrac x a}\right|$.

That also corresponds to what I see when I draw the graph of the integrand and see it is always positive.

So how do we reconcile this fact that it's obviously positive with my evaluation where it comes out negative?

Prime Mover
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  • I'm sorry, but when you integrate for negative $x$, the variable moves in the negative direction, so that the definite integral of a positive function is negative. – Bernard Sep 26 '20 at 21:03
  • I see where you are coming from, but I'm considering the antiderivative here, I'm not doing the definite integral. Of course when you put the limits in on a definite integral it all works out the way you want it to, but for a pure antiderivative where you don't have limits, how does it work? – Prime Mover Sep 26 '20 at 21:25
  • Apply the first fundamental theorem of integral calculus to obtain the antiderivative, and you'll see it works. – Bernard Sep 26 '20 at 21:28

1 Answers1

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You can always use the fundamental theorem of calculus in order to find an antiderivative. So, consider, for $x\in(-\infty,-|a|)$, $$ \int_{-2|a|}^x\dfrac{1}{\sqrt{t^2-a^2}}\,dt $$ With $t=-u$, this becomes $$ -\int_{2|a|}^{-x}\frac{1}{\sqrt{u^2-a^2}}\,du=-\Bigl[\operatorname{arcosh}\frac{u}{a}\Bigr]_{2|a|}^{-x} $$ So the generic antiderivative is $$ -\operatorname{arcosh}\frac{-x}{|a|}+c=-\operatorname{arcosh}\Bigl|\frac{x}{a}\Bigr|+c $$ Indeed, over $(-\infty,-|a|)$, the function is $$ -\operatorname{arcosh}\frac{-x}{|a|} $$ so the chain rule cancels the minusi signs.

You might unify with $$ \operatorname{sgn}x\operatorname{arcosh}\Bigl|\frac{x}{a}\Bigr|+c $$ (of course the constant may be different on the different connected components).

A further check: granted that the derivative of $f(t)=\operatorname{arcosh}t$ is $f'(t)=1/\sqrt{t^2-1}$ and the function is defined for $t\ge1$, but not differentiable at $1$, the chain rule applied to $$ g(x)=\operatorname{arcosh}\Bigl|\frac{x}{a}\Bigr|=\operatorname{arcosh}\frac{|x|}{|a|} $$ yields $$ g'(x)=\frac{1}{|a|}\frac{|x|}{x}\frac{1}{\sqrt{(x/a)^2-1}}=\frac{|x|}{x}\frac{1}{\sqrt{x^2-a^2}} $$

egreg
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  • So are you saying that for negative $x$ the integral genuinely is the minus of the arcosh? My working is correct and the book is wrong? – Prime Mover Sep 26 '20 at 21:28
  • @PrimeMover Yes, indeed. The derivative of $|x|$ is $|x|/x$, if you want to apply the chain rule. I'll add it. – egreg Sep 26 '20 at 21:31
  • Thank you, I will add a note to my webpage on this subject that the source work is flawed. Sorted. – Prime Mover Sep 26 '20 at 21:40
  • Isn't the name of the function $\operatorname{argcosh}$? – Bernard Sep 26 '20 at 21:40
  • @Bernard There are several conventions. Here “ar” stands for “area”. Of course $\cosh^{-1}$ is mathematically wrong. – egreg Sep 26 '20 at 22:27
  • Yes, it's a computer science notation. As a student, I learnt ‘arg’ for ‘argument’. – Bernard Sep 26 '20 at 22:31
  • @Bernard Actually, as far as I know, it's also the ISO notation. I don't disagree with ISO on every point they make. – egreg Sep 26 '20 at 22:41
  • @egreg Saying that $cosh^{-1}$ is "wrong" is a bit dogmatic. It's used plenty. You just have to bear in mind that it can lead to misinterpretation of the superscript ${}^{-1}$ as a power rather than as a function inversion. – Prime Mover Sep 27 '20 at 07:11
  • @PrimeMover The “hyperbolic cosine” function is not invertible. A restriction thereof is, of course, but it's not the same function. This is why I say “mathematically wrong”. – egreg Sep 27 '20 at 07:48
  • @egreg And again, I refuse to accept the concept of "wrong" when it comes to notation. Notation means what you want it to mean. – Prime Mover Sep 27 '20 at 10:42