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Evaluate $\int_\gamma\frac{f'(z)}{f(z)}\ dz$ where $\gamma$ is the line segment $[-a,a]$.

It looks like I could just parametrize so that it becomes
$$I = \int_{-a}^a \frac{f'(t)}{f(t)}\ dt $$ but then I realise that I can't use the fact that it's the derivative of the logarithm, since the contour seems to pass through $0$, which is a branch point of every branch of log...

What do I do here? I was never taught what happens if there was an actual discontinuity on a contour. Do I just ignore this or what happens?

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    $\begingroup$ You did not state any assumptions on $f$. You probably need $f(z)\neq0$ for all $z\in\gamma$. If that is the case, then I don't see such a problem, because $\int_\gamma f'(z)/f(z){\rm d}z=\int_\gamma{\rm d}\ln(f(z))=\ln(f(a))-\ln(f(-a))$. $\endgroup$ – Gerhard S. Nov 8 '17 at 9:46
  • $\begingroup$ I had just thought of that up so I'm not sure of any appropriate assumptions. Maybe that $f$ is holomorphic everywhere in $\mathbb{C}$. I'm unsure because originally I thought I'd have to pick a branch of $\ln$ to use. What if the contour was the line segment $[a,b]$ instead, with $f(a) = 0$ (so the branch point of log is involved)? $\endgroup$ – Twenty-six colours Nov 8 '17 at 9:49
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    $\begingroup$ $$\int_{-\text{a}}^\text{a}\frac{\text{f}\space'\left(x\right)}{\text{f}\left(x\right)}\space\text{d}x=\ln\left|\text{f}\left(\text{a}\right)\right|-\ln\left|\text{f}\left(-\text{a}\right)\right|=\ln\left|\frac{\text{f}\left(\text{a}\right)}{\text{f}\left(-\text{a}\right)}\right|$$ $\endgroup$ – Jan Nov 8 '17 at 10:51
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You are afraid because $\gamma$ goes through $0$ but that is not important. You have to check if the image of $t\mapsto \frac{f'(\gamma(t))}{f(\gamma(t))}$ is in $\mathbb C^-:=\{z\in \mathbb C~:~\Im z\neq 0\vee\Re z>0\}$ and is welldefined. In that case $g:z\mapsto \frac1z$ is the derivative of $\log$ and for $\gamma:[a,b]\to\mathbb C$ you get $$ \int_\gamma\frac{f'(z)}{f(z)}~dz=\int_\gamma (\log\circ f)'(z)~dz=\log(f(\gamma(b)))-\log(f(\gamma(a))). $$ If that is not possible, i.e. if $\gamma$ is a closed curve, than you need either a nice theorem or you have to compute it by hand, using the definition of $f$ and known real integrals for the real and imaginary part.

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