1
$\begingroup$

Suppose that $f$ is analytic on and outside the simple closed negatively oriented contour $\Gamma$. Assume further that $f$ is analytic at $\infty$ and $f(\infty) = 0$. Prove that $f(z)=\frac{1}{2\pi i}\oint_\Gamma \frac{f(\xi)}{\xi -z}d\xi$.

My attempt:

Denote $C$ is the positive oriented circle containing $\Gamma$. Since $f$ is analytic on and outside $\Gamma$, then $\oint_C\frac{f(\xi)}{\xi-z}d\xi+\oint_\Gamma\frac{f(\xi)}{\xi- z}d\xi=0$. So $\oint_\Gamma\frac{f(\xi)}{\xi - z}d\xi=-\oint_C \frac{f(\xi)}{\xi-z}d\xi$. But I have no idea what is the next step. I think maybe this is related to Cauchy's integral formula.

$\endgroup$
0
$\begingroup$

What you have is not right. I assume $z$ is between $\Gamma$ and $C$, in which case you cannot use Cauchy's theorem, but you use Cauchy integral formula (what's under the integral is not analytic between the two curves). So your integrals differ by the value of the function at $z$ (times $2\pi i$ of course). Then you can approximate the integral over $C$ and let the radius go to infinity to find that the integral over $C$ goes to zero. This is where you use that $f$ is analytic at infinity and $f(\infty) =0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.