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Suppose $\varphi$ is a continuous complex valued function on a closed contour $\gamma$. Let $$F(z)=\int_{\gamma} \frac{\varphi(w)}{w-z}~dw$$ Show $\lim_{z\to\infty} F(z)=0$

We may bound:

$$\begin{align*}|F(z)|&\le \int_{\gamma} \left|\frac{\varphi(w)}{w-z}\right|~dw\\&\le M\int_{\gamma}\frac{1}{|w-z|}~dw\end{align*}$$

Then as $z\to\infty$, $\frac{1}{|w-x|}\to 0$ so $F(z)\to 0$.

Is this correct? Seems too trivial.

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    I'd say the bound $\;M\;$ depends on $\;\gamma\;$ and thus it could have an effect on that limit... – 2017-02-27
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    This is indeed correct. – 2017-02-27
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    @DonAntonio But $\gamma$ is fixed – 2017-02-27
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    @ZacharySelk Yep, that the variable of integration is $\;w\;$ misled me. Then yes: I think your proof is correct. – 2017-02-27
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    for $w \in \gamma$ : $\frac{1}{w-z} \to 0$ uniformly and so $\int_\gamma \frac{\varphi(w)}{w-z} dw \to 0$ – 2017-02-27

1 Answers 1

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$$ \begin{align*} |F(z)|&\le \int_{\gamma} \left|\frac{\varphi(w)}{w-z}\right|~dw\\&\le M\int_{\gamma}\frac{1}{|z-w|}~dw \le\frac{M\,\mathrm{length}(\gamma)}{\mathrm{dist}(z,\gamma)} =\frac{c}{\mathrm{dist}(z,\gamma)}. \end{align*} $$ So, if $\gamma\subset D(0,R)$, for some $R>0$, then $$ |F(z)|\le \frac{c}{|z|-R}\to 0, $$
as $z\to\infty$.