Using complex number in integration

Question

I know,

$$I_1 = \int \dfrac{dx}{\sqrt{x^2 - 1}} = \ln|x + \sqrt{x^2-1}| + c$$

But if I factor out $i$ from the denominator, I get:

$$I_2 = -i \int \dfrac{dx}{\sqrt{1 - x^2}} = -i \sin^{-1}x + ic$$

Are these 2 expressions equivalent?

Answer 1

Here you're starting to get into complex territory. Specifically, the $\arcsin(x)$ function for reals is defined on $|x|<1$, but your $\ln$ expression has a $\sqrt{x^2-1}$ which is defined on $|x|>1$, exactly where the normal arcsine isn't defined for real values.

Is this the end of the story? No, but you need to tread with care.

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Let's say you want to make sense of $\ln(z)$ where $z$ is complex, such that $\ln(x + \sqrt{x^2 -1})$ makes sense on $|x|<1$ same as the real $\arcsin(x)$ function. Well, whenever you have a logarithmic expression like $\ln(\text{Something})$, what that expression means is that it's some value such that if you apply it to the exponential function you get the something back. In other words: $$e^{\ln(\text{Something})} = \text{Something}$$ So what happens if we pick $\text{Something}$ to be a complex number (which is what we're looking for)? If $\text{Something} =i$ then $\ln(i)$ should be a value such that $$ e^{\color{blue}{\ln(i)}} = i $$ And here, by remembering Euler's formula $e^{\theta i} = \cos(\theta) + i \sin(\theta)$ we see that $e^{\color{blue}{\frac{\pi}{2}i}} = \cos\left(\frac{\pi}{2}\right)+ i \sin\left(\frac{\pi}{2}\right) = i$. So one reasonable conclusion could be that $$ \color{blue}{\ln(i)} = \color{blue}{ \frac{\pi}{2}i} $$ But there's a problem we glossed over. Since $\sin$ and $\cos$ are periodic, $i \frac{\pi}{2}$ is not the only value that'll work. Indeed, we see that $$ i = e^{\color{blue}{\frac{-3\pi}{2}i}} = e^{\color{blue}{i\frac{5\pi}{2}}} = e^{\color{blue}{\frac{-7\pi}{2}i}}=e^{\color{blue}{\frac{9\pi}{2}i}} = e^{\color{blue}{\frac{-11\pi}{2}i}} = \dots $$ And now we have an infinite possibility for values of $\ln(i)$. Here's where the idea of branch cuts comes into play, where what you do by "picking a branch" is essentially selecting one value for $\ln(i)$ out of the infinite list we showed above and we work with that as its value.

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So now, let's say you pick the branch cut such that $$ \ln(i) = \frac{\pi}{2}i $$ This choice is usually called the "principal branch". With this choice fixed, we can even make more substantial conclusions. Notice that $$ \ln\left(x + \sqrt{x^2-1}\right) +C \Bigg\vert_{x=0} = C+\ln(i) = C+\frac{\pi}{2}i $$ And also $$ -i \arcsin(x)\Bigg\vert_{x=0} =-i(0) =0 $$ So if we want these functions to be equal we must have $C+\frac{\pi}{2}i = 0 \implies C = -\frac{\pi}{2}i$, and thus, you can assert that $$ \boxed{-i\arcsin(x) = \ln\left(x + \sqrt{x^2-1}\right)-\frac{\pi}{2}i, \qquad |x|<1} $$ with $\ln(x)$ being defined on the principal branch. It's important to note that if we chose a different branch, then the constant added would need to be different, so the above equality only holds for the particular branch choice we made, but not for other choices.

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tl;dr there is a way in which we can say that these functions can be equal, but one needs to be careful since when dealing with complex inputs these functions become multi-valued, so equalities need to be handled with care for them to hold any meaning.

Answer 2

Let $\sin^{-1}x = θ$ $$ \sin θ = x\\ \frac{e^{iθ} - e^{-iθ}}{2i} = x \\ $$

Solving the quadratic,

$$ e^{iθ} = ix \mp \sqrt{1 - x^2} = i(x \mp \sqrt{x^2 - 1})\\ $$

Taking log on both sides and then multiplying the numerator and denominator by the congugate,

$$\begin{align} iθ &= \ln i + \ln |x \mp \sqrt{x^2 - 1}| \\ i \sin^{-1}x &= \ln e^{\frac{iπ}2} + \ln |\frac 1{x \pm \sqrt{x^2 - 1}}|\\ &= \frac{iπ}2 - \ln |x \pm \sqrt{x^2 - 1}|\\ \end{align}$$

So $ - i \sin^{-1}x$ is equivalent to $\ln |x ± \sqrt{x^2 - 1}|$.