Does $\int_{[1,z]}\frac{1}{u}du=\log(z)$ where $z\in\mathbb C$?

Question

Does $$\int_{[1,z]}\frac{1}{u}du=\log(z)$$ where $z\in\mathbb C$ ? I know that on a closed circle that contain $0$ we have $$\int_C\frac{1}{z}dz=2i\pi=\log(1),$$

but for $$\int_{[1,z]}\frac{1}{u}du=\log(z)$$ I don't really know to compute the integral.

Answer 1

If $\log(z)$ is the main branch of the logarithm, then, yes, it is true. Of course, I am assuming that $[1,z]$ is that path $t\mapsto 1+t(z-1)$ ($[t\in[0,1]$).

Let $L(z)=\int_{[1,z]}\frac1u\,\mathrm du$. Then $L$ is differentiable and $L'(z)=\frac1z$. This is so because, by Morera's theorem,$$L(z+h)-L(z)=\int_{[1,z+h]}\frac1u\,\mathrm du-\int_{[1,z]}\frac1u\,\mathrm du=\int_{z,z+h}\frac1u\,\mathrm du$$and therefore\begin{align}\lim_{h\to0}\frac{L(z+h)-L(z)-\frac hz}h&=\lim_{h\to0}\frac{\int_{[z,z+h]}\frac1u\,\mathrm du-\frac hz}h\\&=\lim_{h\to0}\frac{\int_{z,z+h}\frac1u-\frac1z\,\mathrm dz}h.\end{align}Now, take $\varepsilon>0$. Fix a $\delta>0$ such that $|w-z|<\delta\implies\left|\frac1w-\frac1z\right|<\varepsilon$. Then, if $|h|<\delta$,$$\left|\frac{\int_{z,z+h}\frac1u-\frac1z\,\mathrm dz}h\right|\leqslant\frac{|h|\varepsilon}{|h|}=\varepsilon.$$So, this proves that$$\lim_{h\to0}\frac{L(z+h)-L(z)-\frac hz}h=0,$$which means that $L'(z)=\frac1z$.

Now, I will prove that, for each $z\in\mathbb{C}\setminus(-\infty,0]$, $L(z)$ is a logarithm of $z$. Let $g(z)=\frac{e^{L(z)}}z$, Then$$g'(z)=\frac{z\frac1ze^{L(z)}-e^{L(z)}}{z^2}=0.$$So, $g$ is constant. Since $g(1)=\frac{e^0}1=1$, $g=1$ and therefore $e^{L(z)}=z$. In other words, $L(z)$ is a logarithm of $z$.

It remains to be proved that $L(z)=\log(z)$. Let $h(z)=L(z)-\log(z)$. Then$$(\forall z\in\mathbb{C}\setminus(-\infty,0]):h(z)\in2\pi i z,$$since, for each $z$, $h(z)$ is the difference between two logarithms of $z$. But $h$ is continuous and its domain is connected. Since the only connected and non-empty subsets of $2\pi i\mathbb Z$ are those with a single element and since $h(1)=0$, then $h\left(\mathbb{C}\setminus(-\infty,0]\right)=\{0\}$. In other words, $h$ is the null function. So, $L=\log$.

Answer 2

Your equation, as written, is meaningless. The integral $$ \int_{[1,z]}\frac{1}{u}\mathrm du $$ has no unique value: it depends on how many times the path $\gamma(0)=1,\gamma(1)=z$ winds around the origin. On the other hand, the logarithm $$ \log(z) $$ has no unique value: the equation $\mathrm e^w=z$ has an infinite number of solutions, $w\in\log|z|+i (\arg z)\pi\mathbb Z$, with (say) $-\pi<\arg z\le \pi$.

If we agree to regard the integral as a symbolic notation for all the possible values it takes as $\gamma$ ranges over $\pi_1(\mathbb C\setminus\{0\})\cong\mathbb Z$, and the logarithm as the set of all solutions of $\mathrm e^w=z$, then both sides are identical, as sets (i.e., they contain the same elements).

Answer 3

If $z$ is not a nonpositive real (i.e. $z$ is not real, or $z$ is real and $z>0$), and $[1,z]$ is the segment from $1$ to $z$, then $\int_{[1,z]} \frac{du}{u}=\mathrm{Log}(z)$, the principal logarithm. This is defined by the property that it agrees with the real logarithm on the positive real axis. Equivalently, it is defined by the branch cut through the negative real axis and the convention $\arg(z) \in (-\pi,\pi)$.