Inverse of Integration?

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Question:

In my homework problem, I'm trying to get rid of an integration over a close curve on the complex plane. I was wondering if its possible to somehow take the inverse of the integral on other side of the following equatin.

Please explain the reasoning of why or why not so that I can understand it for future problems :P.

2ipif'(0) = Integral over alpha (Any circle containing 0 on the interior) of f(x)/x^2.

Where f(x) is an analytic function from c to c.

Obviously f(x)/x^2 is not analytic on all c, since 0 is a critical point, I was still wondering if there is a way to get rid of the integral by doing something on both sides of the equation. Thanks for any help you can provide.
 

Answers and Replies

  • #2
Gib Z
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Does the integral have bounds? If so, no inverse. Otherwise, differentiate.
 
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D H
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Use the residue theorem.
 
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HallsofIvy
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Question:

In my homework problem, I'm trying to get rid of an integration over a close curve on the complex plane. I was wondering if its possible to somehow take the inverse of the integral on other side of the following equatin.

Please explain the reasoning of why or why not so that I can understand it for future problems :P.

2ipif'(0) = Integral over alpha (Any circle containing 0 on the interior) of f(x)/x^2.

Where f(x) is an analytic function from c to c.

Obviously f(x)/x^2 is not analytic on all c, since 0 is a critical point, I was still wondering if there is a way to get rid of the integral by doing something on both sides of the equation. Thanks for any help you can provide.
What you have is a special case of Cauchy's integral formula:
[tex]2\pi i f^{(n)}(z_0)= \int \frac{f(z)}{(z-z_0)^{n+1}}dz[/tex]
where the integration is over any closed path (not necessarily a circle) with z0[/sup] in its interior and f(z) is analytic inside the path. You can prove it by writing f as a Taylor's series, dividing each term by (z- z0)n+1 and then integrating term by term. It is easy to show that [itex]\int (z-z_0)^n dz[/itex] over a closed path containing z0 is 0 for any n other than -1 and is [itex]2\pi i[/itex] when n= -1.
 

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