Question on integrating $\frac{1}{2\pi}\int_0^{2\pi} \log(e^{ix})dx$?

Question

I evaluated the integral $$ \frac{1}{2\pi}\int_0^{2\pi} \log(e^{ix})dx, $$ in Wolfram Alpha and the result was $0$. But if I do it analytically I get $$ \frac{1}{2\pi}\int_0^{2\pi} \log(e^{ix})dx = \frac{1}{2\pi}\int_0^{2\pi} ix \ dx = i 2 \pi. $$

So why does Wolfram Alpha say that the integral is $0$?

Answer 1

Wolfram Alpha computes Log function which returns the principal value of $\log(r e^{i\theta}) = \log(r) + \phi$, where $\phi \in (-\pi,\pi]$ is the principal argument (for example, if $\theta = 3\pi/2$ then $\phi=-\pi/2$). Basically, you are evaluating $$\frac{i}{2\pi} \int_{-\pi}^{\pi} x dx.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {1 \over 2\pi}\int_{0}^{2\pi}\ln\pars{\expo{\ic x}}\,\dd x & = {1 \over 2\pi}\oint_{\verts{z} = 1}\ln\pars{z}\,{\dd z \over \ic z}\qquad \pars{~\substack{\ds{\ln\ \mbox{is its}\ Principal\ Branch.}\\[3mm] \mbox{Integration is performed along a}\\[1mm] \ds{key-hole\ contour}\ \mbox{which 'takes care'} \\[1mm] \mbox{of the}\ \ds{\ln}\ \mbox{branch-cut.}}~} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}\,\,\,& -\,{1 \over 2\pi\ic}\int_{-1}^{-\epsilon}{\ln\pars{-x} + \ic\pi\over x}\,\dd x - {1 \over 2\pi\ic}\int_{\pi}^{-\pi}{\ln\pars{\epsilon} + \ic\theta \over \epsilon\expo{\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta \\ & -\,{1 \over 2\pi\ic}\int_{-\epsilon}^{-1}{\ln\pars{-x} - \ic\pi\over x}\,\dd x \\[5mm] & = {1 \over 2\pi\ic}\int_{\epsilon}^{1}{\ln\pars{x} + \ic\pi\over x}\,\dd x + \ln\pars{\epsilon} - {1 \over 2\pi\ic}\int_{\epsilon}^{1}{\ln\pars{x} - \ic\pi\over x}\,\dd x \\[5mm] & = -\,{1 \over 2}\,\ln\pars{\epsilon} + \ln\pars{\epsilon} - {1 \over 2}\,\ln\pars{\epsilon} \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\large\to}\,\,\, \bbx{0} \end{align}