Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was studying calculus 1 other day, in the subject of integral of partial fractions, when I perceived that the well know indefinite integral: $$ \int \dfrac{dx}{1+x^2} $$ could be done another way, just by factoring it in $\mathbb{C}[x]$. So I did $1+x^2=(x+i)(x-i)$. Hence $$ \dfrac{1}{1+x^2}=\dfrac{A(x-i)+B(x+i)}{1+x^2} $$ for some A, B constants. It is easy to see that $A=i/2$ and $B=-i/2$. Then $$ \int \dfrac{dx}{1+x^2}=\int \left(\dfrac{i/2}{x+i}+\dfrac{-i/2}{x-i} \right) dx = \dfrac{i}{2}\ln{|x+i|}-\dfrac{i}{2}\ln{|x-i|}+c=$$ $$=\dfrac{i}{2}\ln{\left|\dfrac{x+i}{x-i}\right|}+c$$ Finally, my question is: Am I allowed to factorize a polynomial in $\mathbb{C}[x]$ to integrate it, and if I am, what is the relation between the real indefinite integral $\tan^{-1}$ and $\dfrac{i}{2}\ln{\left|\dfrac{x+i}{x-i}\right|}$. Thank you in advance.

share|improve this question
Have you seen this? BTW: if you're dealing with complex numbers in your partial fractions, you really can't use $\ln\,|x|$ as an antiderivative anymore, and some more care is needed. –  Guess who it is. Jan 26 '12 at 22:37
@J.M. Hmm. I think this question is useless as I could googled it and found an easy answer. Thanks. –  joaopaulolf Jan 26 '12 at 22:47
If you have found an easy answer (and if it adds anything to what has already been posted here), you should post it, or a link to it, here. –  Gerry Myerson Jan 26 '12 at 22:49

1 Answer 1

up vote 1 down vote accepted

What you've done is correct as far as it goes. The big complication, which I think might just be the only reason why complex numbers are usually avoided in first-year calculus courses, is that the logarithm ($\ln$ or $\log$) function and the $\arctan$ function are multiple-valued. The multiple-valued nature of the arctangent is mentioned in the trigometry course you took before taking calculus; that of the logarithm is apparent when you learn that $e^{i\theta} = \cos\theta + i\sin\theta$, recallying that $\cos$ and $\sin$ are not one-to-one. Once you've seen that exponential functions are trigonometric functions, it might not be too surprising that logarithmic functions are inverse-trigonometric functions.

Later addition: If you write $$ \begin{align} \cos x & = \frac{e^{ix}+e^{-ix}}{2} \\ \\ \\ \sin x & = \frac{e^{ix}-e^{-ix}}{2i} \end{align} $$ you can then find $y=\tan x$ as a function of $e^{ix}$. Multiplying the numerator and denominator by $e^{ix}$ you get an expression in which $e^{2ix}$ occurse twice, and with further algebra, you can solve for $e^{2ix}$ and then for $x$ as a function of $y$, if you allow multiple-valued inverse functions.

The conventional way of defining the concept of "function" for the past century or so rules out "multiple-valued" functions, so any use of that term will cause some mathematicians some discomfort.

share|improve this answer
as J. M. noted above, the absolute value in the argument of the logarithm is incorrect. –  Fabian Jan 26 '12 at 22:43
Thank you. Now that you mentioned this, I can see the relationship, and it's not complicated how I thought it would be. –  joaopaulolf Jan 26 '12 at 22:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.