%path = "maths/functions/integral of 1÷z" %kind = kinda["texts"] %level = 12
From real calculus we know, that \(\frac{dy}{dx}=\frac{d\,e^x}{dx}=e^x=y\), and therefore for the inverse \(\ln\) \(\frac{dx}{dy}=\frac{d\,\ln y}{dy}=\frac{1}{y}\) for \(y>0\). For the antiderivative of \(\frac{1}{y}\) we can include negative \(y\), if we take the absolute value: \(\int\frac{1}{y}dy=ln|y|+C\). This follows from the symmetry of \(\frac{1}{y}\). At 0 there is a singularity, i.e. one cannot integrate over it.
In \(\mathbb{C}\) we have \(e^z=e^{x+iy}=e^xe^{iy}\), i.e. the real part becomes the absolute value \(e^x\) and the imaginary part becomes the argument. That is the reason for the period \(2\pi i\) along the imaginary axis. The antiderivative of \(\frac{1}{z}\) is the inverse of \(e^z\), which means that the absolute value becomes the real part \(ln|z|\) and the argument becomes the imaginary part.
\(\int \frac{1}{z}dz=ln|z|+i\arg(z)+C\)
In \(\mathbb{C}\) one can integrate around the singularity:
This is the precursor of the residue theorem.