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.

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    $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Jan 26 '12 at 22:37
  • $\begingroup$ @J.M. Hmm. I think this question is useless as I could googled it and found an easy answer. Thanks. $\endgroup$ – joaopaulolf Jan 26 '12 at 22:47
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jan 26 '12 at 22:49

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.

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

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