Integral of Complex Numbers

Question

I was wondering about doing integration the unorthodox way. For example I took up $$\int\sqrt{1-x^2}\,\mathrm dx$$ and instead of substituting $x$ for $\sin t$ I tried doing it for $\sec t$ which gives me integral of $$\int i\sec t\,\tan^2t\,\mathrm dt$$ I was wondering what's wrong in doing this if we integrate for area in some imaginary dimension. Can you guys guide me to integrating in complex?

Background : I have no knowledge about complex analysis but looking forward to learn.

Answer 1

There's nothing "wrong" with it, but you have to be careful. Some functions that are well-defined over the real numbers become multivalued in the complex numbers.

A simpler example:

$$\int \frac{1}{x^2+1} \; dx = \int \frac{1}{2i} \left(\frac{1}{x-i}-\frac{1}{x+i} \right) \; dx$$

$$=\frac{1}{2i} \left( \ln(x-i)-\ln{x+i} \right)+C$$

$$=\frac{1}{2i}\ln\frac{x-i}{x+i} +C.$$

This is correct but: 1. in the complex numbers $\ln z$ is multivalued so which branch of $\ln z$ are we using here? and 2. the normal answer is $\tan x +C$ and how do we see that these two answers are really the same?

Most universities offer an undergrad course is "complex variables" where such issues are studied. The classic text is Churchill.

Edit: You're asking a lot here. Please google any terms you don't know: Euler's formula gives $e^{it} = \cos t + i\sin t$. By plugging in $t+2\pi$ for $t$ we see that the function $e^z$ has period $2\pi i$. That is $e^{z+2k\pi i} = e^{z}$ for any integer $k$. So if $y=e^{z} = e^{z+2k\pi i}$, when we try to take the inverse function we try to write $\ln y = z +2k\pi i$ (which is a mistake, until we decide what $k$ is. Wiki has a page with nice pictures:

https://en.wikipedia.org/wiki/Complex_logarithm

Answer 2

You are right. You can do that substitution.

Now, if $t$ is real, then $|\sec t| \ge 1$ so we are in cases where $1-x^2 \le 0$ and it is not surprising we get an imaginary answer for $\int\sqrt{1-x^2}\;dx$.

added

Real scenario $$ \int_a^b\sqrt{1-x^2}\;dx\quad\text{is real if } -1\le a < b \le 1 $$ Imaginary scenario $$ \int_a^b\sqrt{1-x^2}\;dx\quad\text{is imaginary if } 1\le a < b $$