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?

PS: I'm in $12$ grade and have no knowledge about complex analysis but looking forward to learn. I don't know MathJax and I'm learning it so please ignore my errors. Thanks.

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    $\begingroup$ There's nothing, strictly speaking, that's wrong with that you've written. However, that integral is much more difficult to compute than the integral of $\cos^2(t)$, and back-substituting to get the answer in terms of $x$ will give you something messier in the method you've chosen. $\endgroup$ – Michael Lee Jul 28 '17 at 20:11
  • $\begingroup$ I understand the difficulty. But what I mean is that if definite integral is area then should it not remain same even during integrating in complex. I tried setting limits myself to check if the value of the computed integrals was same but it wasn't. I intend to know what's wrong... $\endgroup$ – Sajid Rizvi Jul 28 '17 at 20:14
  • $\begingroup$ You will get the same answer because when you perform a change of variables, you change the limits of your integral as well (integrating in the complex plane requires defining a contour, of course, so you'll have to be careful about this). $\endgroup$ – Michael Lee Jul 28 '17 at 20:16

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:


  • $\begingroup$ Brown/Churchill is okay. There are other options, though. I like Ahlfors, for example, as well as Rudin. $\endgroup$ – Michael Lee Jul 28 '17 at 20:18
  • $\begingroup$ Woahhh! Could you please elaborate about the multivalued part? Also does definite integral in complex mean area? $\endgroup$ – Sajid Rizvi Jul 28 '17 at 20:19
  • $\begingroup$ The solution to $e^x = y$ over $\mathbb{C}$ is $x = \log(r)+i(\theta+2k\pi)$, where $y = re^{i\theta}$ and $k\in \mathbb{Z}$. This defines the multivalued complex logarithm $\log(y)$. We might write the principal value as $\operatorname{Log}(y)$. $\endgroup$ – Michael Lee Jul 28 '17 at 20:22
  • $\begingroup$ See edit, but you've touched on a big question. $\endgroup$ – B. Goddard Jul 28 '17 at 20:27
  • $\begingroup$ I get it now. I thought about this a few days back like log(-i) to the base i can be any number of the form 4k+3 where k belongs to N $\endgroup$ – Sajid Rizvi Jul 28 '17 at 20:34

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 $\int\sqrt{1-x^2}\;dx$.


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

  • $\begingroup$ If I set limits on the integral then that would refer to area under the curve but why does it come out to be different in the real and complex scenario $\endgroup$ – Sajid Rizvi Jul 28 '17 at 20:15

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