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I want to evaluate the following integral $\int_{\gamma} 1/z \, dz$ where $\gamma$ is the semi-circle in the upper half plane from $1$ to $-1$.

Can I just use the Fundamental Theorem of Calculus here? I know that if $\gamma$ crosses the negative real axis I can't but what about here where it touches it?

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    I would recommend computing the integral (Not a difficult calculation), then look at the value obtained from FTC, then asking yourself these questions. – 2017-01-15

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Why not directly as line integral?

$$z=e^{it}\;,\;\;t\in [0,\pi]\implies dz=ie^{it}dt\implies$$

and now :

$$\int_\gamma\frac{dz}z=\int_0^\pi\frac{i\,e^{it}}{e^{it}}dt=\pi i$$

If you insist in using the FTC: take Log$\,z\;$, but remove the ray $\;iy\;,\;\;y\in(-\infty,0]\;$ , thus taking the argument of a number in the remaining domain to be an angle in $\;\left(-\frac\pi2,\,\frac{3\pi}2\right)\;$ , and thus:

$$\left.\int_\gamma\frac{dz}z=\text{Log}\,z\right|_1^{-1}=\text{Log}\,(-1)-\text{Log}\,1=\require{cancel}\cancel{\log1}+i\arg(-1)-(\cancel{\log1}-i\arg 1)=i\pi+i\cdot0$$

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    I know that it's a simple calculation as a line integral, I was just interested to know whether or not I could use the FTC instead. Can I? – 2017-01-15
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    @Si.0788 Read the added stuff. – 2017-01-15
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    So as long as the curve doesn't enclose the origin this can always be done? – 2017-01-15
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    @Si.0788 If it is a simple curve, yes: I think so. – 2017-01-15