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Use the principal branch of log z $\int_{-1}^{1} log z\, dz$

My attempt was we see how -1 to 1 became pi to 0? That should make the e^whatever terms go away. Then we can do it by parts. $\int_{-1}^{1} log z\, dz$ =$\int_{\pi}^{0} log (e^{i\theta})ie^{i\theta}\, dz=-2+i\pi$

Is this correct result? Could please show me another method of resolution?

Could someone help me through this problem?

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1 Answer 1

up vote 2 down vote accepted

I would say $$ \int_{-1}^1 \log(z)\,dz = \int_0^1\log(x)\,dx+\int_{-1}^0[\log(-x)+i\pi]\,dx $$ and then do some real integrals.

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