Integration a long closed curve is 0

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SUMMARY

The discussion centers on the evaluation of complex integrals around closed curves, specifically addressing the integral \(\int_{\gamma (t)} f(z) dz\). It is established that if the function \(f\) is analytic within the region enclosed by the curve \(\gamma\), then the integral evaluates to zero, as indicated by the Cauchy integral theorem. The participants also discuss the relationship between analytic functions and conservative functions, emphasizing that the absence of singularities within the closed path is crucial for the integral's evaluation.

PREREQUISITES
  • Understanding of complex analysis concepts, particularly analytic functions.
  • Familiarity with the Cauchy integral theorem and Cauchy-Riemann equations.
  • Knowledge of contour integration techniques.
  • Basic understanding of conservative vector fields in the context of complex functions.
NEXT STEPS
  • Study the Cauchy integral theorem in detail to understand its implications for complex integrals.
  • Learn about the Cauchy-Riemann equations and their role in defining analytic functions.
  • Investigate the properties of conservative functions and their mathematical definitions.
  • Explore examples of contour integration and practice evaluating integrals around closed curves.
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in advanced calculus, particularly those studying integrals in complex domains.

alyafey22
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\int_{\gamma (t) }\, f(z) dz

\int_{\alpha}^{\beta} \, f(\gamma (t))\, \gamma '(t) \, dt

\text{Use the substitution : } \gamma (t) = \xi

\int_{\gamma (\alpha) }^{\gamma (\beta)} \, f(\xi )\, d \xi

\text{If we integrate around a closed loope : }\gamma (\alpha) = \gamma(\beta)

\int_{\gamma (\alpha) }^{\gamma (\alpha)} \, f( \xi )\, d \xi =0

\text{This is only true if the function is analytic }

Feel free to leave any comments .
 
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ZaidAlyafey said:
\int_{\gamma (t) }\, f(z) dz

\int_{\alpha}^{\beta} \, f(\gamma (t))\, \gamma '(t) \, dt

\text{Use the substitution : } \gamma (t) = \xi

\int_{\gamma (\alpha) }^{\gamma (\beta)} \, f(\xi )\, d \xi

\text{If we integrate around a closed loope : }\gamma (\alpha) = \gamma(\beta)

\int_{\gamma (\alpha) }^{\gamma (\alpha)} \, f( \xi )\, d \xi =0

\text{This is only true if the function is analytic }

Feel free to leave any comments .
Comment, not a full answer.

This will work if your function f is conservative. I don't how that relates to the analyticity of f.

-Dan
 
topsquark said:
Comment, not a full answer.

This will work if your function f is conservative. I don't how that relates to the analyticity of f.

-Dan

But how to define conservative functions mathematically ?
We define analytic functions as those which satisfy the Cauchy-Reimann equations and the partial derivatives exist and are continuous so if they have a pole then we can use the Cauchy-integral formula to find the integral along the loop this is illustrated by the deformation hypothesis .
 
ZaidAlyafey said:
But how to define conservative functions mathematically ?
We define analytic functions as those which satisfy the Cauchy-Reimann equations and the partial derivatives exist and are continuous so if they have a pole then we can use the Cauchy-integral formula to find the integral along the loop this is illustrated by the deformation hypothesis .
Ah! It's a complex integration. You didn't tell us that. (Tmi)

Then as far as I know, so long as you have a closed path (that doesn't contain any nasty singularities) then the answer is 0.

-Dan

Come to think about it, if it's analytic I think that means no singularities. I'm too lazy to check that. Time for a nap! (Yawn)

-Dan
 
If $f$ is analytic inside the region enclosed by $\gamma$, the integral in question will be zero. Are you putting forth a proof of that? I'm a little unclear what it is you're after.
 
Ackbach said:
If $f$ is analytic inside the region enclosed by $\gamma$, the integral in question will be zero. Are you putting forth a proof of that? I'm a little unclear what it is you're after.

Yes, indeed.
 
What are your assumptions? What theorems are you allowed to use?
 
I am using the contour integration formual :

\int_{\gamma (t) }\, f(z) dz =\int_{\alpha}^{\beta} \, f(\gamma (t))\, \gamma '(t) \, dt
 

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