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Let $f(z) := \frac{\mathbb{e}^{iz}}{z}$ $z \in \mathbb{C}$ where $0 \notin \mathbb{C}$

I need to show that when $C_k$, a semi circle of radius e is traversed in the clockwise direction is traversed in the clockwise direction $ \int_{C_k} f(z) dz \to -i\pi $ as $ k\to 0$

So firstly i set up a parameterization of the semi circle $z(t) = e^{-it}$ where $-\pi\leq t \leq 0 $ (I'm not sure this is totally correct) I then change the variable to get the integral $\int_{C_k} f(z) dz$ = $\int_{-\pi}^0 -ie^{ike^{-it}} dt$

I then try and bound this using the fact that $| \int_{C_k} f(z) dz | \leq \int_{C_k} |f(z)| |dz|$ but i cannot seem to find a bound!

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  • $\begingroup$ First expand $\frac{e^{iz}}{z}$ in a Laurent series centered at $z=0$. $\endgroup$ Commented Dec 2, 2013 at 6:39
  • $\begingroup$ Write $\large z \not= 0$ instead of $\large 0 \not\in {\mathbb C}$. $\endgroup$ Commented Dec 2, 2013 at 7:20
  • $\begingroup$ Check this ---> math.stackexchange.com/a/589173/85343 $\endgroup$ Commented Dec 2, 2013 at 7:21
  • $\begingroup$ Thanks Felix, but i cannot see how to explicitly solve my problem. $\endgroup$
    – John wayne
    Commented Dec 2, 2013 at 7:35

1 Answer 1

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Write $$f(z) = \frac{1}{z} + \frac{e^{iz}-1}{z}$$ The integral of $1/z$ of the half-circle is $-\pi i $ by direct computation. The second term has a removable singularity at $0$: $$\frac{e^{iz}-1}{z} \to i ,\quad z\to 0$$ Hence, it is bounded in a neighborhood of $0$. Say, $$\left|\frac{e^{iz}-1}{z}\right|\le M \quad \text{when } |z|<r_0$$ Integrating this over a half-circle of radius $r$ gives at most $M\pi r$, which tends to zero as $r\to0$.

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